Introduction to harmonic analysis

Sasha Sodin and Matthias Taufer

March 16 2020

Lecture notes for an LTCC course FebruaryndashMarch 2020 Thesections marked with an asterisk were not covered in class Pleasedrop a note (at the email address below) if you spot any mistakes

Contents1 Introduction 2

2 Fourier analysis on ZNZ 221 Discrete Fourier transform 222 Rothrsquos theorem 423 Fast Fourier transform 7

3 Fourier series 931 Fourier coefficients and Plancherel theorem 932 Convergence of Fourier series 1233 Fejer summation 16

4 Fourier series further topics and applications 1841 Polynomial approximation 1842 The heat equation 2343 Toeplitz matrices and the Szego theorem 2644 Central limit theorem 3245 Equidistribution modulo one 35

5 Fourier transform 3951 Introduction 3952 Fourier transform in Schwartz space 4053 Poisson summation 4254 The uncertainty principle 44

Some notation 46

References 47School of Mathematical Sciences Queen Mary University of London London E1 4NS United Kingdom E-mailasodin mtaeuferqmulacuk

1

1 Introduction

Harmonic analysis is used to study problems possessing translation invarianceFor example it can be used to diagonalise operators commuting with shifts

Example 11 For any r0 r1 r2 isin C the matrix

A = A[r0 r1 r2] =⎛⎜⎝

r0 r1 r2

r2 r0 r1

r1 r2 r0

⎞⎟⎠

is invariant under simultaneous cyclic shifts of the rows and the columns in otherwords it commutes with the shifts A[010] and A[001] Let ω = exp(2πi3)and

e0 =⎡⎢⎢⎢⎢⎢⎣

111

⎤⎥⎥⎥⎥⎥⎦ e1 =

⎡⎢⎢⎢⎢⎢⎣

1ωω2

⎤⎥⎥⎥⎥⎥⎦ e2 =

⎡⎢⎢⎢⎢⎢⎣

1ω2

ω

⎤⎥⎥⎥⎥⎥⎦

Then

Ae0 = (r0 + r1 + r2)e0 Ae1 = (r0 + r1ω + r2ω2)e1 Ae2 = (r0 + r1ω

2 + r2ω)e2

Remarkably the eigenvectors do not depend on r0 r1 r21

There are many other operators thatr commute with shifts eg

d

dxd2

dx2 f ↦ f lowast g et cet

and there are many problems with (explicit or hidden) translation invarianceThis will be the scope of the applications that we shall consider in this minicourse

Among the numerous books on Fourier analysis we have been inspired by thatof Dym and McKean [1972] which highlights the role of Fourier transform asa versatile tool in applications of various kinds We have also borrowed somematerial from the monographs of Montgomery [2014] Katznelson [2004] Korner[1989] Stein and Shakarchi [2003] Lanczos [1966] similarly to the last two ofthese we do not rely on Lebesgue integration

2 Fourier analysis on ZNZ

21 Discrete Fourier transform

Equip the space of functions f ∶ ZNZ rarr C ie of N -periodic functions from Zto C with the inner product

⟨f g⟩ =Nminus1

sumk=0

f(k)g(k)

1Remarkably but not surprisingly all the matrices A commute and in fact form a commutative algebra

2

This will be our inner product space `2(ZNZ) Let

ep(k) = exp(2πipkN) p k isin ZNZ

Claim 21

1 ep ∶ ZNZ are (the) characters of ZNZ ie ep(k + `) = ep(k)ep(`)

2 1radicNep form an orthonormal basis of `2(ZNZ)

Proof Item 1 is obvious item 2 is verified directly

1

N⟨ep eq⟩ =

1

N

Nminus1

sumk=0

exp(2πik(p minus q)N) =⎧⎪⎪⎨⎪⎪⎩

1 p equiv q mod N

0

Consider the (left) shift S ∶ `2(ZNZ)rarr `2(ZNZ) (Sf)(k) = f(k + 1) Fromitem 1

(Sep)(k) = ep(k + 1) = ep(1)ep(k) ie Sep = ep(1)ep

Theorem 22 If A ∶ `2(ZNZ)rarr `2(ZNZ) is a linear map that commutes withS ie AS = SA then A is diagonal in the basis (ep)

Proof We have SAep = ASep = ep(1)Aep hence Aep is an eigenvector of S witheigenvalue ep(1) ie a multiple λpep of ep

Exercise 23 Let A ∶ `2(ZNZ)rarr `2(ZNZ) be a linear map

1 The map A commutes with S iff it is a convolution operator ie thereexists r ∶ ZNZrarr C such that

(Af)(k) = (r lowast f)(k) =Nminus1

sum`=0

r(k minus `)f(`) (21)

2 If A is of the form (21) the eigenvalue corresponding to ep equals

λp =Nminus1

sumk=0

r(k)ep(k) = ⟨r ep⟩ = r(p)

in the notation of

Definition 24 The Fourier coefficients of f isin `2(ZNZ) are the numbers

f(p) = ⟨f ep⟩ =Nminus1

sumk=0

f(k) exp(minus2πipkN)

the map F ∶ f ↦ f is called the discrete Fourier transform

3

Remark 25 If we let L2(ZNZ) be the space of N -periodic functions equippedwith the inner product

⟪φψ⟫ = 1

N

Nminus1

sump=0

φ(p)ψ(p)

then F ∶ `2(ZNZ) rarr L2(ZNZ) is an isometry To keep notation consistentwe shall denote functions of k by Roman letters f g⋯ and functions of p ndash byGreek letters φψ⋯

Remark 26 By definition f(p) measures the correlation between f and ep In-formally this coefficient is responsible for the correlation between f and patternsof period q where pq equiv 1 modulo N

Exercise 27 The discrete Fourier transform boasts the following properties

1 the inverse transform Flowast ∶ L2(ZNZ)rarr `2(ZNZ) is given by

Flowast ∶ φ↦ φ φ(k) = 1

N

Nminus1

sumk=0

φ(p)ek(p)

2 Let (f lowast g)(k) = sum f(k minus `)g(`) (φ ψ)(p) = 1N sumφ(p minus q)ψ(q) Then

F(f lowast g) = F(f)F(g) and F(fg) = F(f)F(g)

3 For f equiv 1N one has f(p) = Nδp0 and for g(k) = δk0 one has g(p) equiv 1

22 Rothrsquos theorem

In this section we describe the following somewhat sophisticated application

Theorem 28 (Roth [1952]) For any δ isin (01] there exists N0(δ) such that forany N ge N0(δ) any A sub 1⋯N of cardinality ∣A∣ ge δN contains a three-termarithmetic progression

See Gowers [1998] and Soundararajan [2010] for a survey of the field includingthe more recent developments

Proof The property of being an arithmetic progression is translation invariantTherefore it is not unexpected that Fourier analysis can be used However a fewpreliminary reductions have to be made For convenience denote the assertionof the theorem (ldquothere exists N(δ) arithmetic progressionrdquo) by Rothδ

First Rothδ is clearly true for δ gt 23 (in this case A if large enough containsthree consecutive integers) Therefore it suffices to show that

Rothδ(1+δ25)OcircrArr Rothδ

4

Second it suffices to consider prime N Formally if for a certain δprime gt 0 anda certain N prime

0(δprime) one has that for any prime N prime ge N prime0(δ) any set A sub 1⋯N prime

of cardinality ∣A∣ ge δprimeN contains a three-term arithmetic progression then Rothδholds for all δ gt δprime Indeed by Bertrandrsquos postulate there is a prime

lfloor(δ minus δprime)N2rfloor le N prime le 2lfloor(δ minus δprime)N2rfloor

Partitioning 1⋯N primelfloorNN primerfloor into lfloorNN primerfloor chunks Cj of size N prime we find that forleast one of the chunks ∣A cap Cj ∣ ge δprimeN prime and hence A cap Cj contains a three-termarithmetic progression

Third to make use of the Fourier transform we shall work modulo N Torule out ldquowraparoundrdquo (ie to make sure that the three-term progression we willhave found is not say N minus 4N minus 12) we consider several cases If

∣a isin A ∶ a le N3∣ gt δ(1 + δ20)∣A∣3 (22)

the assumption Rothδ(1+δ25) ensures that there is an arithmetic progression inA1 = a isin A ∶le N1 = lfloorN3rfloor We can argue similarly if

∣a isin A ∶ a gt 2N3∣ gt δ(1 + δ20)∣A∣3 (23)

Therefore we assume for the rest of the proof that (22) and (23) fail ie that

B = a isin A ∶ N3 lt a le 2N3

is of cardinality ∣B∣ ge δ(1minus δ10)∣A∣3 Observe that if a isin A and b c isin B are suchthat a + c equiv 2b mod N then either (a b c) is a proper arithmetic progression(without wraparound) or a = b = c

Now the Fourier-analytic part begins The triples (a b c) with a + c equiv 2bmod N can be counted as follows

S = sumaisinAbcisinB

1a+c=2b =N

sumabc=1

1A(a)1B(b)1B(c)1a+c=2b (24)

= 1

N

N

sumabc=1

Nminus1

sump=0

1A(a)1B(b)1B(c) exp(minus2πip(a minus 2b + c)N) (25)

= 1

N

Nminus1

sump=0

1A(p)1B(minus2p)1B(p) (26)

Observing that 1A(0) = ∣A∣ and similarly for BC and using the CauchyndashSchwarz

5

inequality and that N is odd we get

S ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(minus2p)1B(p)∣ (27)

ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(p)∣2 (28)

= 1

N∣A∣ ∣B∣2 minusmax

pne0∣1A(p)∣ ∣B∣ (29)

This may be understood as follows Among the sim N arithmetic progressions(a b c) sim ∣A∣N have a isin A and similarly sim ∣B∣N have b isin B or c isin B If theseevents would be independent (which is the case if A is random) we would haveS sim N 3 times (∣A∣N) times (∣B∣N)2 = ∣A∣∣B∣2N If the non-zero Fourier coefficients of1A are small A has no distinguished patterns and therefore we expect this tobe a good approximation

Proceeding with the argument consider two cases If maxpne0 ∣1A(p)∣ le δ∣A∣6then

(29) ge ∣A∣∣B∣(∣B∣N minus δ6) ge ∣A∣∣B∣(δ(1 minus δ10)3 minus δ6) ge c(δ)N 2

which is much greater than N and we are doneNow assume that there exists p ne 0 such that ∣1A(p)∣ gt δ∣A∣6 (Recall that this

means that A is correlated with some pattern of period p) In this case we shalldecompose 1⋯N into several arithmetic progressions so that the density ofA in at least one of them is at least δ(1 + δ50) Formally let m = lceil100

δ rceil and let

Pj = 0 le x le N minus 1 ∶ jNm

le px lt (j + 1)Nm

0 le j ltm

These are arithmetic progressions modulo N indeed if r be such that pr equiv 1mod N then

Pj = kr ∶ jNm

le k lt (j + 1)Nm

Then

∣1A(p)∣ = ∣sumjsumxisinPj

1A(x) exp(minus2πipxN)∣ (210)

le ∣sumjsumxisinPj

1A(x) exp(minus2πijm)∣ + 2π

msumjsumxisinPj

1A(x) (211)

=RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRR+ 2π∣A∣

m (212)

6

Hence RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRRge ∣1A(p)∣ minus

2π∣A∣m

ge δ∣A∣12

(213)

If all the sets AcapPj were of the same size the sum would vanish Quantitativelywe have

Exercise 29 Prove if f ∶ ZmZrarr R and f = 1msum

mminus1j=0 f(j) then

mminus1maxj=0

f(j) ge f + 1

2m∣mminus1

sumj=0

f(j) exp(minus2πijm)∣ (214)

Applying (214) to f(j) = ∣A cap Pj ∣ we obtain that there exists j such that

∣A cap Pj ∣ ge∣A∣m

+ δ∣A∣24m

ge δ(1 + δ25)∣Pj ∣

Now we further decompose this Pj into proper arithmetic progressions Among

the lfloorradicNrfloor numbers r2r⋯ lfloor

radicNrfloorr there are two which are not far from one

another rlowast = k2r minus k1r mod N isin 1⋯ lfloorradicNrfloor Let klowast = k2 minus k1 note that

klowast leradicN For each 1 le k le klowast split Pj into klowast arithmetic progressions of step

rlowast (still possibly with wraparound) and then split each of these into proper

arithmetic progressions without wraparound All but at most CradicN of these

progressions have length ge cradicN hence at least one of them P prime

j sub Pj satisfies

∣P primej ∣ ge c

radicN ∣AcapP prime

j ∣ ge δ(1+δ25)∣P primej ∣ By the induction assumption AcapP prime

j containsa three-term arithmetic progression

Exercise 210 Show that

1 N0(δ) le exp exp(Cδ)

2 the assumption ∣A∣ ge δN can be replaced with ∣A∣ ge C1N log logN for asufficiently large constant C1

23 Fast Fourier transform

The naıve implementation of the discrete Fourier transform on ZNZ requiressim constN 2 arithmetic operations Fast Fourier transform allows to do the jobwith O(N logN) operations See Montgomery [2014] for a historical discussion

Denote the smallest number of arithmetic operations required for the compu-tation of the Fourier transform by n(N)

Proposition 211 n(2N) le 2n(N) + 3N

7

Proof Let f ∶ Z2NZrarr C Construct two auxiliary functions g h ∶ ZNZrarr C

g(n) = f(2n) h(n) = f(2n + 1)

Then

f(p) =2Nminus1

sumn=0

f(n) exp(minus2πinp(2N))

=Nminus1

sumn=0

f(2n) exp(minus2πi2np(2N)) +Nminus1

sumn=0

f(2n + 1) exp(minus2πi (2n + 1)p(2N))

= g(p) + exp(minusπipN)h(p)

which means that for 0 le p le N minus 1

f(p) = g(p)+ exp(minusπipN)h(p) f(p+N) = g(p)minus exp(minusπipN)h(p) (215)

The computation of f from g and h using (215) requires 3N arithmetic opera-tions (if we assume that the values of exp(minusπipN) can be precomputed)

Corollary 212 n(2n) le (32)n2n for n ge 1

Proof By induction

This corollary implies that n(N) le CN logN when N is a power of 2 Similarreasoning applies to N which do not have large prime factors The general caserequires a separate argument for large prime N which we do not reproduce here

Applications From item 2 of Exercise 27 we obtain that the convolution f lowastg =(f g)or of two functions f g ∶ ZNZ rarr C can be computed in O(n(N)) steps forN = 2n this is O(N logN)

Next p(x) = sumdk=0 akxk and q(x) = sumdk=0 bkx

k be two polynomials Their productis equal to (pq)(x) = sum2d

k=0(sumkl=0 albkminusl)xk This looks quite similar to a convolu-tion except that there is no wraparound

Exercise 213 The product of two polynomials of degree d can be computed inO(d log d) arithmetic operations

Exercise 214 Long multiplication of two d-digit numbers takes O(d2) operations(where an operation means addition or multiplication of single-digit numbers)Propose an algorithm which would only require O(d log10 d) operations

8

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

1 Introduction

Harmonic analysis is used to study problems possessing translation invarianceFor example it can be used to diagonalise operators commuting with shifts

Example 11 For any r0 r1 r2 isin C the matrix

A = A[r0 r1 r2] =⎛⎜⎝

r0 r1 r2

r2 r0 r1

r1 r2 r0

⎞⎟⎠

is invariant under simultaneous cyclic shifts of the rows and the columns in otherwords it commutes with the shifts A[010] and A[001] Let ω = exp(2πi3)and

e0 =⎡⎢⎢⎢⎢⎢⎣

111

⎤⎥⎥⎥⎥⎥⎦ e1 =

⎡⎢⎢⎢⎢⎢⎣

1ωω2

⎤⎥⎥⎥⎥⎥⎦ e2 =

⎡⎢⎢⎢⎢⎢⎣

1ω2

ω

⎤⎥⎥⎥⎥⎥⎦

Then

Ae0 = (r0 + r1 + r2)e0 Ae1 = (r0 + r1ω + r2ω2)e1 Ae2 = (r0 + r1ω

2 + r2ω)e2

Remarkably the eigenvectors do not depend on r0 r1 r21

There are many other operators thatr commute with shifts eg

d

dxd2

dx2 f ↦ f lowast g et cet

and there are many problems with (explicit or hidden) translation invarianceThis will be the scope of the applications that we shall consider in this minicourse

Among the numerous books on Fourier analysis we have been inspired by thatof Dym and McKean [1972] which highlights the role of Fourier transform asa versatile tool in applications of various kinds We have also borrowed somematerial from the monographs of Montgomery [2014] Katznelson [2004] Korner[1989] Stein and Shakarchi [2003] Lanczos [1966] similarly to the last two ofthese we do not rely on Lebesgue integration

2 Fourier analysis on ZNZ

21 Discrete Fourier transform

Equip the space of functions f ∶ ZNZ rarr C ie of N -periodic functions from Zto C with the inner product

⟨f g⟩ =Nminus1

sumk=0

f(k)g(k)

1Remarkably but not surprisingly all the matrices A commute and in fact form a commutative algebra

2

This will be our inner product space `2(ZNZ) Let

ep(k) = exp(2πipkN) p k isin ZNZ

Claim 21

1 ep ∶ ZNZ are (the) characters of ZNZ ie ep(k + `) = ep(k)ep(`)

2 1radicNep form an orthonormal basis of `2(ZNZ)

Proof Item 1 is obvious item 2 is verified directly

1

N⟨ep eq⟩ =

1

N

Nminus1

sumk=0

exp(2πik(p minus q)N) =⎧⎪⎪⎨⎪⎪⎩

1 p equiv q mod N

0

Consider the (left) shift S ∶ `2(ZNZ)rarr `2(ZNZ) (Sf)(k) = f(k + 1) Fromitem 1

(Sep)(k) = ep(k + 1) = ep(1)ep(k) ie Sep = ep(1)ep

Theorem 22 If A ∶ `2(ZNZ)rarr `2(ZNZ) is a linear map that commutes withS ie AS = SA then A is diagonal in the basis (ep)

Proof We have SAep = ASep = ep(1)Aep hence Aep is an eigenvector of S witheigenvalue ep(1) ie a multiple λpep of ep

Exercise 23 Let A ∶ `2(ZNZ)rarr `2(ZNZ) be a linear map

1 The map A commutes with S iff it is a convolution operator ie thereexists r ∶ ZNZrarr C such that

(Af)(k) = (r lowast f)(k) =Nminus1

sum`=0

r(k minus `)f(`) (21)

2 If A is of the form (21) the eigenvalue corresponding to ep equals

λp =Nminus1

sumk=0

r(k)ep(k) = ⟨r ep⟩ = r(p)

in the notation of

Definition 24 The Fourier coefficients of f isin `2(ZNZ) are the numbers

f(p) = ⟨f ep⟩ =Nminus1

sumk=0

f(k) exp(minus2πipkN)

the map F ∶ f ↦ f is called the discrete Fourier transform

3

Remark 25 If we let L2(ZNZ) be the space of N -periodic functions equippedwith the inner product

⟪φψ⟫ = 1

N

Nminus1

sump=0

φ(p)ψ(p)

then F ∶ `2(ZNZ) rarr L2(ZNZ) is an isometry To keep notation consistentwe shall denote functions of k by Roman letters f g⋯ and functions of p ndash byGreek letters φψ⋯

Remark 26 By definition f(p) measures the correlation between f and ep In-formally this coefficient is responsible for the correlation between f and patternsof period q where pq equiv 1 modulo N

Exercise 27 The discrete Fourier transform boasts the following properties

1 the inverse transform Flowast ∶ L2(ZNZ)rarr `2(ZNZ) is given by

Flowast ∶ φ↦ φ φ(k) = 1

N

Nminus1

sumk=0

φ(p)ek(p)

2 Let (f lowast g)(k) = sum f(k minus `)g(`) (φ ψ)(p) = 1N sumφ(p minus q)ψ(q) Then

F(f lowast g) = F(f)F(g) and F(fg) = F(f)F(g)

3 For f equiv 1N one has f(p) = Nδp0 and for g(k) = δk0 one has g(p) equiv 1

22 Rothrsquos theorem

In this section we describe the following somewhat sophisticated application

Theorem 28 (Roth [1952]) For any δ isin (01] there exists N0(δ) such that forany N ge N0(δ) any A sub 1⋯N of cardinality ∣A∣ ge δN contains a three-termarithmetic progression

See Gowers [1998] and Soundararajan [2010] for a survey of the field includingthe more recent developments

Proof The property of being an arithmetic progression is translation invariantTherefore it is not unexpected that Fourier analysis can be used However a fewpreliminary reductions have to be made For convenience denote the assertionof the theorem (ldquothere exists N(δ) arithmetic progressionrdquo) by Rothδ

First Rothδ is clearly true for δ gt 23 (in this case A if large enough containsthree consecutive integers) Therefore it suffices to show that

Rothδ(1+δ25)OcircrArr Rothδ

4

Second it suffices to consider prime N Formally if for a certain δprime gt 0 anda certain N prime

0(δprime) one has that for any prime N prime ge N prime0(δ) any set A sub 1⋯N prime

of cardinality ∣A∣ ge δprimeN contains a three-term arithmetic progression then Rothδholds for all δ gt δprime Indeed by Bertrandrsquos postulate there is a prime

lfloor(δ minus δprime)N2rfloor le N prime le 2lfloor(δ minus δprime)N2rfloor

Partitioning 1⋯N primelfloorNN primerfloor into lfloorNN primerfloor chunks Cj of size N prime we find that forleast one of the chunks ∣A cap Cj ∣ ge δprimeN prime and hence A cap Cj contains a three-termarithmetic progression

Third to make use of the Fourier transform we shall work modulo N Torule out ldquowraparoundrdquo (ie to make sure that the three-term progression we willhave found is not say N minus 4N minus 12) we consider several cases If

∣a isin A ∶ a le N3∣ gt δ(1 + δ20)∣A∣3 (22)

the assumption Rothδ(1+δ25) ensures that there is an arithmetic progression inA1 = a isin A ∶le N1 = lfloorN3rfloor We can argue similarly if

∣a isin A ∶ a gt 2N3∣ gt δ(1 + δ20)∣A∣3 (23)

Therefore we assume for the rest of the proof that (22) and (23) fail ie that

B = a isin A ∶ N3 lt a le 2N3

is of cardinality ∣B∣ ge δ(1minus δ10)∣A∣3 Observe that if a isin A and b c isin B are suchthat a + c equiv 2b mod N then either (a b c) is a proper arithmetic progression(without wraparound) or a = b = c

Now the Fourier-analytic part begins The triples (a b c) with a + c equiv 2bmod N can be counted as follows

S = sumaisinAbcisinB

1a+c=2b =N

sumabc=1

1A(a)1B(b)1B(c)1a+c=2b (24)

= 1

N

N

sumabc=1

Nminus1

sump=0

1A(a)1B(b)1B(c) exp(minus2πip(a minus 2b + c)N) (25)

= 1

N

Nminus1

sump=0

1A(p)1B(minus2p)1B(p) (26)

Observing that 1A(0) = ∣A∣ and similarly for BC and using the CauchyndashSchwarz

5

inequality and that N is odd we get

S ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(minus2p)1B(p)∣ (27)

ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(p)∣2 (28)

= 1

N∣A∣ ∣B∣2 minusmax

pne0∣1A(p)∣ ∣B∣ (29)

This may be understood as follows Among the sim N arithmetic progressions(a b c) sim ∣A∣N have a isin A and similarly sim ∣B∣N have b isin B or c isin B If theseevents would be independent (which is the case if A is random) we would haveS sim N 3 times (∣A∣N) times (∣B∣N)2 = ∣A∣∣B∣2N If the non-zero Fourier coefficients of1A are small A has no distinguished patterns and therefore we expect this tobe a good approximation

Proceeding with the argument consider two cases If maxpne0 ∣1A(p)∣ le δ∣A∣6then

(29) ge ∣A∣∣B∣(∣B∣N minus δ6) ge ∣A∣∣B∣(δ(1 minus δ10)3 minus δ6) ge c(δ)N 2

which is much greater than N and we are doneNow assume that there exists p ne 0 such that ∣1A(p)∣ gt δ∣A∣6 (Recall that this

means that A is correlated with some pattern of period p) In this case we shalldecompose 1⋯N into several arithmetic progressions so that the density ofA in at least one of them is at least δ(1 + δ50) Formally let m = lceil100

δ rceil and let

Pj = 0 le x le N minus 1 ∶ jNm

le px lt (j + 1)Nm

0 le j ltm

These are arithmetic progressions modulo N indeed if r be such that pr equiv 1mod N then

Pj = kr ∶ jNm

le k lt (j + 1)Nm

Then

∣1A(p)∣ = ∣sumjsumxisinPj

1A(x) exp(minus2πipxN)∣ (210)

le ∣sumjsumxisinPj

1A(x) exp(minus2πijm)∣ + 2π

msumjsumxisinPj

1A(x) (211)

=RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRR+ 2π∣A∣

m (212)

6

Hence RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRRge ∣1A(p)∣ minus

2π∣A∣m

ge δ∣A∣12

(213)

If all the sets AcapPj were of the same size the sum would vanish Quantitativelywe have

Exercise 29 Prove if f ∶ ZmZrarr R and f = 1msum

mminus1j=0 f(j) then

mminus1maxj=0

f(j) ge f + 1

2m∣mminus1

sumj=0

f(j) exp(minus2πijm)∣ (214)

Applying (214) to f(j) = ∣A cap Pj ∣ we obtain that there exists j such that

∣A cap Pj ∣ ge∣A∣m

+ δ∣A∣24m

ge δ(1 + δ25)∣Pj ∣

Now we further decompose this Pj into proper arithmetic progressions Among

the lfloorradicNrfloor numbers r2r⋯ lfloor

radicNrfloorr there are two which are not far from one

another rlowast = k2r minus k1r mod N isin 1⋯ lfloorradicNrfloor Let klowast = k2 minus k1 note that

klowast leradicN For each 1 le k le klowast split Pj into klowast arithmetic progressions of step

rlowast (still possibly with wraparound) and then split each of these into proper

arithmetic progressions without wraparound All but at most CradicN of these

progressions have length ge cradicN hence at least one of them P prime

j sub Pj satisfies

∣P primej ∣ ge c

radicN ∣AcapP prime

j ∣ ge δ(1+δ25)∣P primej ∣ By the induction assumption AcapP prime

j containsa three-term arithmetic progression

Exercise 210 Show that

1 N0(δ) le exp exp(Cδ)

2 the assumption ∣A∣ ge δN can be replaced with ∣A∣ ge C1N log logN for asufficiently large constant C1

23 Fast Fourier transform

The naıve implementation of the discrete Fourier transform on ZNZ requiressim constN 2 arithmetic operations Fast Fourier transform allows to do the jobwith O(N logN) operations See Montgomery [2014] for a historical discussion

Denote the smallest number of arithmetic operations required for the compu-tation of the Fourier transform by n(N)

Proposition 211 n(2N) le 2n(N) + 3N

7

Proof Let f ∶ Z2NZrarr C Construct two auxiliary functions g h ∶ ZNZrarr C

g(n) = f(2n) h(n) = f(2n + 1)

Then

f(p) =2Nminus1

sumn=0

f(n) exp(minus2πinp(2N))

=Nminus1

sumn=0

f(2n) exp(minus2πi2np(2N)) +Nminus1

sumn=0

f(2n + 1) exp(minus2πi (2n + 1)p(2N))

= g(p) + exp(minusπipN)h(p)

which means that for 0 le p le N minus 1

f(p) = g(p)+ exp(minusπipN)h(p) f(p+N) = g(p)minus exp(minusπipN)h(p) (215)

The computation of f from g and h using (215) requires 3N arithmetic opera-tions (if we assume that the values of exp(minusπipN) can be precomputed)

Corollary 212 n(2n) le (32)n2n for n ge 1

Proof By induction

This corollary implies that n(N) le CN logN when N is a power of 2 Similarreasoning applies to N which do not have large prime factors The general caserequires a separate argument for large prime N which we do not reproduce here

Applications From item 2 of Exercise 27 we obtain that the convolution f lowastg =(f g)or of two functions f g ∶ ZNZ rarr C can be computed in O(n(N)) steps forN = 2n this is O(N logN)

Next p(x) = sumdk=0 akxk and q(x) = sumdk=0 bkx

k be two polynomials Their productis equal to (pq)(x) = sum2d

k=0(sumkl=0 albkminusl)xk This looks quite similar to a convolu-tion except that there is no wraparound

Exercise 213 The product of two polynomials of degree d can be computed inO(d log d) arithmetic operations

Exercise 214 Long multiplication of two d-digit numbers takes O(d2) operations(where an operation means addition or multiplication of single-digit numbers)Propose an algorithm which would only require O(d log10 d) operations

8

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

This will be our inner product space `2(ZNZ) Let

ep(k) = exp(2πipkN) p k isin ZNZ

Claim 21

1 ep ∶ ZNZ are (the) characters of ZNZ ie ep(k + `) = ep(k)ep(`)

2 1radicNep form an orthonormal basis of `2(ZNZ)

Proof Item 1 is obvious item 2 is verified directly

1

N⟨ep eq⟩ =

1

N

Nminus1

sumk=0

exp(2πik(p minus q)N) =⎧⎪⎪⎨⎪⎪⎩

1 p equiv q mod N

0

Consider the (left) shift S ∶ `2(ZNZ)rarr `2(ZNZ) (Sf)(k) = f(k + 1) Fromitem 1

(Sep)(k) = ep(k + 1) = ep(1)ep(k) ie Sep = ep(1)ep

Theorem 22 If A ∶ `2(ZNZ)rarr `2(ZNZ) is a linear map that commutes withS ie AS = SA then A is diagonal in the basis (ep)

Proof We have SAep = ASep = ep(1)Aep hence Aep is an eigenvector of S witheigenvalue ep(1) ie a multiple λpep of ep

Exercise 23 Let A ∶ `2(ZNZ)rarr `2(ZNZ) be a linear map

1 The map A commutes with S iff it is a convolution operator ie thereexists r ∶ ZNZrarr C such that

(Af)(k) = (r lowast f)(k) =Nminus1

sum`=0

r(k minus `)f(`) (21)

2 If A is of the form (21) the eigenvalue corresponding to ep equals

λp =Nminus1

sumk=0

r(k)ep(k) = ⟨r ep⟩ = r(p)

in the notation of

Definition 24 The Fourier coefficients of f isin `2(ZNZ) are the numbers

f(p) = ⟨f ep⟩ =Nminus1

sumk=0

f(k) exp(minus2πipkN)

the map F ∶ f ↦ f is called the discrete Fourier transform

3

Remark 25 If we let L2(ZNZ) be the space of N -periodic functions equippedwith the inner product

⟪φψ⟫ = 1

N

Nminus1

sump=0

φ(p)ψ(p)

then F ∶ `2(ZNZ) rarr L2(ZNZ) is an isometry To keep notation consistentwe shall denote functions of k by Roman letters f g⋯ and functions of p ndash byGreek letters φψ⋯

Remark 26 By definition f(p) measures the correlation between f and ep In-formally this coefficient is responsible for the correlation between f and patternsof period q where pq equiv 1 modulo N

Exercise 27 The discrete Fourier transform boasts the following properties

1 the inverse transform Flowast ∶ L2(ZNZ)rarr `2(ZNZ) is given by

Flowast ∶ φ↦ φ φ(k) = 1

N

Nminus1

sumk=0

φ(p)ek(p)

2 Let (f lowast g)(k) = sum f(k minus `)g(`) (φ ψ)(p) = 1N sumφ(p minus q)ψ(q) Then

F(f lowast g) = F(f)F(g) and F(fg) = F(f)F(g)

3 For f equiv 1N one has f(p) = Nδp0 and for g(k) = δk0 one has g(p) equiv 1

22 Rothrsquos theorem

In this section we describe the following somewhat sophisticated application

Theorem 28 (Roth [1952]) For any δ isin (01] there exists N0(δ) such that forany N ge N0(δ) any A sub 1⋯N of cardinality ∣A∣ ge δN contains a three-termarithmetic progression

See Gowers [1998] and Soundararajan [2010] for a survey of the field includingthe more recent developments

Proof The property of being an arithmetic progression is translation invariantTherefore it is not unexpected that Fourier analysis can be used However a fewpreliminary reductions have to be made For convenience denote the assertionof the theorem (ldquothere exists N(δ) arithmetic progressionrdquo) by Rothδ

First Rothδ is clearly true for δ gt 23 (in this case A if large enough containsthree consecutive integers) Therefore it suffices to show that

Rothδ(1+δ25)OcircrArr Rothδ

4

Second it suffices to consider prime N Formally if for a certain δprime gt 0 anda certain N prime

0(δprime) one has that for any prime N prime ge N prime0(δ) any set A sub 1⋯N prime

of cardinality ∣A∣ ge δprimeN contains a three-term arithmetic progression then Rothδholds for all δ gt δprime Indeed by Bertrandrsquos postulate there is a prime

lfloor(δ minus δprime)N2rfloor le N prime le 2lfloor(δ minus δprime)N2rfloor

Partitioning 1⋯N primelfloorNN primerfloor into lfloorNN primerfloor chunks Cj of size N prime we find that forleast one of the chunks ∣A cap Cj ∣ ge δprimeN prime and hence A cap Cj contains a three-termarithmetic progression

Third to make use of the Fourier transform we shall work modulo N Torule out ldquowraparoundrdquo (ie to make sure that the three-term progression we willhave found is not say N minus 4N minus 12) we consider several cases If

∣a isin A ∶ a le N3∣ gt δ(1 + δ20)∣A∣3 (22)

the assumption Rothδ(1+δ25) ensures that there is an arithmetic progression inA1 = a isin A ∶le N1 = lfloorN3rfloor We can argue similarly if

∣a isin A ∶ a gt 2N3∣ gt δ(1 + δ20)∣A∣3 (23)

Therefore we assume for the rest of the proof that (22) and (23) fail ie that

B = a isin A ∶ N3 lt a le 2N3

is of cardinality ∣B∣ ge δ(1minus δ10)∣A∣3 Observe that if a isin A and b c isin B are suchthat a + c equiv 2b mod N then either (a b c) is a proper arithmetic progression(without wraparound) or a = b = c

Now the Fourier-analytic part begins The triples (a b c) with a + c equiv 2bmod N can be counted as follows

S = sumaisinAbcisinB

1a+c=2b =N

sumabc=1

1A(a)1B(b)1B(c)1a+c=2b (24)

= 1

N

N

sumabc=1

Nminus1

sump=0

1A(a)1B(b)1B(c) exp(minus2πip(a minus 2b + c)N) (25)

= 1

N

Nminus1

sump=0

1A(p)1B(minus2p)1B(p) (26)

Observing that 1A(0) = ∣A∣ and similarly for BC and using the CauchyndashSchwarz

5

inequality and that N is odd we get

S ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(minus2p)1B(p)∣ (27)

ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(p)∣2 (28)

= 1

N∣A∣ ∣B∣2 minusmax

pne0∣1A(p)∣ ∣B∣ (29)

This may be understood as follows Among the sim N arithmetic progressions(a b c) sim ∣A∣N have a isin A and similarly sim ∣B∣N have b isin B or c isin B If theseevents would be independent (which is the case if A is random) we would haveS sim N 3 times (∣A∣N) times (∣B∣N)2 = ∣A∣∣B∣2N If the non-zero Fourier coefficients of1A are small A has no distinguished patterns and therefore we expect this tobe a good approximation

Proceeding with the argument consider two cases If maxpne0 ∣1A(p)∣ le δ∣A∣6then

(29) ge ∣A∣∣B∣(∣B∣N minus δ6) ge ∣A∣∣B∣(δ(1 minus δ10)3 minus δ6) ge c(δ)N 2

which is much greater than N and we are doneNow assume that there exists p ne 0 such that ∣1A(p)∣ gt δ∣A∣6 (Recall that this

means that A is correlated with some pattern of period p) In this case we shalldecompose 1⋯N into several arithmetic progressions so that the density ofA in at least one of them is at least δ(1 + δ50) Formally let m = lceil100

δ rceil and let

Pj = 0 le x le N minus 1 ∶ jNm

le px lt (j + 1)Nm

0 le j ltm

These are arithmetic progressions modulo N indeed if r be such that pr equiv 1mod N then

Pj = kr ∶ jNm

le k lt (j + 1)Nm

Then

∣1A(p)∣ = ∣sumjsumxisinPj

1A(x) exp(minus2πipxN)∣ (210)

le ∣sumjsumxisinPj

1A(x) exp(minus2πijm)∣ + 2π

msumjsumxisinPj

1A(x) (211)

=RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRR+ 2π∣A∣

m (212)

6

Hence RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRRge ∣1A(p)∣ minus

2π∣A∣m

ge δ∣A∣12

(213)

If all the sets AcapPj were of the same size the sum would vanish Quantitativelywe have

Exercise 29 Prove if f ∶ ZmZrarr R and f = 1msum

mminus1j=0 f(j) then

mminus1maxj=0

f(j) ge f + 1

2m∣mminus1

sumj=0

f(j) exp(minus2πijm)∣ (214)

Applying (214) to f(j) = ∣A cap Pj ∣ we obtain that there exists j such that

∣A cap Pj ∣ ge∣A∣m

+ δ∣A∣24m

ge δ(1 + δ25)∣Pj ∣

Now we further decompose this Pj into proper arithmetic progressions Among

the lfloorradicNrfloor numbers r2r⋯ lfloor

radicNrfloorr there are two which are not far from one

another rlowast = k2r minus k1r mod N isin 1⋯ lfloorradicNrfloor Let klowast = k2 minus k1 note that

klowast leradicN For each 1 le k le klowast split Pj into klowast arithmetic progressions of step

rlowast (still possibly with wraparound) and then split each of these into proper

arithmetic progressions without wraparound All but at most CradicN of these

progressions have length ge cradicN hence at least one of them P prime

j sub Pj satisfies

∣P primej ∣ ge c

radicN ∣AcapP prime

j ∣ ge δ(1+δ25)∣P primej ∣ By the induction assumption AcapP prime

j containsa three-term arithmetic progression

Exercise 210 Show that

1 N0(δ) le exp exp(Cδ)

2 the assumption ∣A∣ ge δN can be replaced with ∣A∣ ge C1N log logN for asufficiently large constant C1

23 Fast Fourier transform

The naıve implementation of the discrete Fourier transform on ZNZ requiressim constN 2 arithmetic operations Fast Fourier transform allows to do the jobwith O(N logN) operations See Montgomery [2014] for a historical discussion

Denote the smallest number of arithmetic operations required for the compu-tation of the Fourier transform by n(N)

Proposition 211 n(2N) le 2n(N) + 3N

7

Proof Let f ∶ Z2NZrarr C Construct two auxiliary functions g h ∶ ZNZrarr C

g(n) = f(2n) h(n) = f(2n + 1)

Then

f(p) =2Nminus1

sumn=0

f(n) exp(minus2πinp(2N))

=Nminus1

sumn=0

f(2n) exp(minus2πi2np(2N)) +Nminus1

sumn=0

f(2n + 1) exp(minus2πi (2n + 1)p(2N))

= g(p) + exp(minusπipN)h(p)

which means that for 0 le p le N minus 1

f(p) = g(p)+ exp(minusπipN)h(p) f(p+N) = g(p)minus exp(minusπipN)h(p) (215)

The computation of f from g and h using (215) requires 3N arithmetic opera-tions (if we assume that the values of exp(minusπipN) can be precomputed)

Corollary 212 n(2n) le (32)n2n for n ge 1

Proof By induction

This corollary implies that n(N) le CN logN when N is a power of 2 Similarreasoning applies to N which do not have large prime factors The general caserequires a separate argument for large prime N which we do not reproduce here

Applications From item 2 of Exercise 27 we obtain that the convolution f lowastg =(f g)or of two functions f g ∶ ZNZ rarr C can be computed in O(n(N)) steps forN = 2n this is O(N logN)

Next p(x) = sumdk=0 akxk and q(x) = sumdk=0 bkx

k be two polynomials Their productis equal to (pq)(x) = sum2d

k=0(sumkl=0 albkminusl)xk This looks quite similar to a convolu-tion except that there is no wraparound

Exercise 213 The product of two polynomials of degree d can be computed inO(d log d) arithmetic operations

Exercise 214 Long multiplication of two d-digit numbers takes O(d2) operations(where an operation means addition or multiplication of single-digit numbers)Propose an algorithm which would only require O(d log10 d) operations

8

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Remark 25 If we let L2(ZNZ) be the space of N -periodic functions equippedwith the inner product

⟪φψ⟫ = 1

N

Nminus1

sump=0

φ(p)ψ(p)

then F ∶ `2(ZNZ) rarr L2(ZNZ) is an isometry To keep notation consistentwe shall denote functions of k by Roman letters f g⋯ and functions of p ndash byGreek letters φψ⋯

Remark 26 By definition f(p) measures the correlation between f and ep In-formally this coefficient is responsible for the correlation between f and patternsof period q where pq equiv 1 modulo N

Exercise 27 The discrete Fourier transform boasts the following properties

1 the inverse transform Flowast ∶ L2(ZNZ)rarr `2(ZNZ) is given by

Flowast ∶ φ↦ φ φ(k) = 1

N

Nminus1

sumk=0

φ(p)ek(p)

2 Let (f lowast g)(k) = sum f(k minus `)g(`) (φ ψ)(p) = 1N sumφ(p minus q)ψ(q) Then

F(f lowast g) = F(f)F(g) and F(fg) = F(f)F(g)

3 For f equiv 1N one has f(p) = Nδp0 and for g(k) = δk0 one has g(p) equiv 1

22 Rothrsquos theorem

In this section we describe the following somewhat sophisticated application

Theorem 28 (Roth [1952]) For any δ isin (01] there exists N0(δ) such that forany N ge N0(δ) any A sub 1⋯N of cardinality ∣A∣ ge δN contains a three-termarithmetic progression

See Gowers [1998] and Soundararajan [2010] for a survey of the field includingthe more recent developments

Proof The property of being an arithmetic progression is translation invariantTherefore it is not unexpected that Fourier analysis can be used However a fewpreliminary reductions have to be made For convenience denote the assertionof the theorem (ldquothere exists N(δ) arithmetic progressionrdquo) by Rothδ

First Rothδ is clearly true for δ gt 23 (in this case A if large enough containsthree consecutive integers) Therefore it suffices to show that

Rothδ(1+δ25)OcircrArr Rothδ

4

Second it suffices to consider prime N Formally if for a certain δprime gt 0 anda certain N prime

0(δprime) one has that for any prime N prime ge N prime0(δ) any set A sub 1⋯N prime

of cardinality ∣A∣ ge δprimeN contains a three-term arithmetic progression then Rothδholds for all δ gt δprime Indeed by Bertrandrsquos postulate there is a prime

lfloor(δ minus δprime)N2rfloor le N prime le 2lfloor(δ minus δprime)N2rfloor

Partitioning 1⋯N primelfloorNN primerfloor into lfloorNN primerfloor chunks Cj of size N prime we find that forleast one of the chunks ∣A cap Cj ∣ ge δprimeN prime and hence A cap Cj contains a three-termarithmetic progression

Third to make use of the Fourier transform we shall work modulo N Torule out ldquowraparoundrdquo (ie to make sure that the three-term progression we willhave found is not say N minus 4N minus 12) we consider several cases If

∣a isin A ∶ a le N3∣ gt δ(1 + δ20)∣A∣3 (22)

the assumption Rothδ(1+δ25) ensures that there is an arithmetic progression inA1 = a isin A ∶le N1 = lfloorN3rfloor We can argue similarly if

∣a isin A ∶ a gt 2N3∣ gt δ(1 + δ20)∣A∣3 (23)

Therefore we assume for the rest of the proof that (22) and (23) fail ie that

B = a isin A ∶ N3 lt a le 2N3

is of cardinality ∣B∣ ge δ(1minus δ10)∣A∣3 Observe that if a isin A and b c isin B are suchthat a + c equiv 2b mod N then either (a b c) is a proper arithmetic progression(without wraparound) or a = b = c

Now the Fourier-analytic part begins The triples (a b c) with a + c equiv 2bmod N can be counted as follows

S = sumaisinAbcisinB

1a+c=2b =N

sumabc=1

1A(a)1B(b)1B(c)1a+c=2b (24)

= 1

N

N

sumabc=1

Nminus1

sump=0

1A(a)1B(b)1B(c) exp(minus2πip(a minus 2b + c)N) (25)

= 1

N

Nminus1

sump=0

1A(p)1B(minus2p)1B(p) (26)

Observing that 1A(0) = ∣A∣ and similarly for BC and using the CauchyndashSchwarz

5

inequality and that N is odd we get

S ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(minus2p)1B(p)∣ (27)

ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(p)∣2 (28)

= 1

N∣A∣ ∣B∣2 minusmax

pne0∣1A(p)∣ ∣B∣ (29)

This may be understood as follows Among the sim N arithmetic progressions(a b c) sim ∣A∣N have a isin A and similarly sim ∣B∣N have b isin B or c isin B If theseevents would be independent (which is the case if A is random) we would haveS sim N 3 times (∣A∣N) times (∣B∣N)2 = ∣A∣∣B∣2N If the non-zero Fourier coefficients of1A are small A has no distinguished patterns and therefore we expect this tobe a good approximation

Proceeding with the argument consider two cases If maxpne0 ∣1A(p)∣ le δ∣A∣6then

(29) ge ∣A∣∣B∣(∣B∣N minus δ6) ge ∣A∣∣B∣(δ(1 minus δ10)3 minus δ6) ge c(δ)N 2

which is much greater than N and we are doneNow assume that there exists p ne 0 such that ∣1A(p)∣ gt δ∣A∣6 (Recall that this

means that A is correlated with some pattern of period p) In this case we shalldecompose 1⋯N into several arithmetic progressions so that the density ofA in at least one of them is at least δ(1 + δ50) Formally let m = lceil100

δ rceil and let

Pj = 0 le x le N minus 1 ∶ jNm

le px lt (j + 1)Nm

0 le j ltm

These are arithmetic progressions modulo N indeed if r be such that pr equiv 1mod N then

Pj = kr ∶ jNm

le k lt (j + 1)Nm

Then

∣1A(p)∣ = ∣sumjsumxisinPj

1A(x) exp(minus2πipxN)∣ (210)

le ∣sumjsumxisinPj

1A(x) exp(minus2πijm)∣ + 2π

msumjsumxisinPj

1A(x) (211)

=RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRR+ 2π∣A∣

m (212)

6

Hence RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRRge ∣1A(p)∣ minus

2π∣A∣m

ge δ∣A∣12

(213)

If all the sets AcapPj were of the same size the sum would vanish Quantitativelywe have

Exercise 29 Prove if f ∶ ZmZrarr R and f = 1msum

mminus1j=0 f(j) then

mminus1maxj=0

f(j) ge f + 1

2m∣mminus1

sumj=0

f(j) exp(minus2πijm)∣ (214)

Applying (214) to f(j) = ∣A cap Pj ∣ we obtain that there exists j such that

∣A cap Pj ∣ ge∣A∣m

+ δ∣A∣24m

ge δ(1 + δ25)∣Pj ∣

Now we further decompose this Pj into proper arithmetic progressions Among

the lfloorradicNrfloor numbers r2r⋯ lfloor

radicNrfloorr there are two which are not far from one

another rlowast = k2r minus k1r mod N isin 1⋯ lfloorradicNrfloor Let klowast = k2 minus k1 note that

klowast leradicN For each 1 le k le klowast split Pj into klowast arithmetic progressions of step

rlowast (still possibly with wraparound) and then split each of these into proper

arithmetic progressions without wraparound All but at most CradicN of these

progressions have length ge cradicN hence at least one of them P prime

j sub Pj satisfies

∣P primej ∣ ge c

radicN ∣AcapP prime

j ∣ ge δ(1+δ25)∣P primej ∣ By the induction assumption AcapP prime

j containsa three-term arithmetic progression

Exercise 210 Show that

1 N0(δ) le exp exp(Cδ)

2 the assumption ∣A∣ ge δN can be replaced with ∣A∣ ge C1N log logN for asufficiently large constant C1

23 Fast Fourier transform

The naıve implementation of the discrete Fourier transform on ZNZ requiressim constN 2 arithmetic operations Fast Fourier transform allows to do the jobwith O(N logN) operations See Montgomery [2014] for a historical discussion

Denote the smallest number of arithmetic operations required for the compu-tation of the Fourier transform by n(N)

Proposition 211 n(2N) le 2n(N) + 3N

7

Proof Let f ∶ Z2NZrarr C Construct two auxiliary functions g h ∶ ZNZrarr C

g(n) = f(2n) h(n) = f(2n + 1)

Then

f(p) =2Nminus1

sumn=0

f(n) exp(minus2πinp(2N))

=Nminus1

sumn=0

f(2n) exp(minus2πi2np(2N)) +Nminus1

sumn=0

f(2n + 1) exp(minus2πi (2n + 1)p(2N))

= g(p) + exp(minusπipN)h(p)

which means that for 0 le p le N minus 1

f(p) = g(p)+ exp(minusπipN)h(p) f(p+N) = g(p)minus exp(minusπipN)h(p) (215)

The computation of f from g and h using (215) requires 3N arithmetic opera-tions (if we assume that the values of exp(minusπipN) can be precomputed)

Corollary 212 n(2n) le (32)n2n for n ge 1

Proof By induction

This corollary implies that n(N) le CN logN when N is a power of 2 Similarreasoning applies to N which do not have large prime factors The general caserequires a separate argument for large prime N which we do not reproduce here

Applications From item 2 of Exercise 27 we obtain that the convolution f lowastg =(f g)or of two functions f g ∶ ZNZ rarr C can be computed in O(n(N)) steps forN = 2n this is O(N logN)

Next p(x) = sumdk=0 akxk and q(x) = sumdk=0 bkx

k be two polynomials Their productis equal to (pq)(x) = sum2d

k=0(sumkl=0 albkminusl)xk This looks quite similar to a convolu-tion except that there is no wraparound

Exercise 213 The product of two polynomials of degree d can be computed inO(d log d) arithmetic operations

Exercise 214 Long multiplication of two d-digit numbers takes O(d2) operations(where an operation means addition or multiplication of single-digit numbers)Propose an algorithm which would only require O(d log10 d) operations

8

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Second it suffices to consider prime N Formally if for a certain δprime gt 0 anda certain N prime

0(δprime) one has that for any prime N prime ge N prime0(δ) any set A sub 1⋯N prime

of cardinality ∣A∣ ge δprimeN contains a three-term arithmetic progression then Rothδholds for all δ gt δprime Indeed by Bertrandrsquos postulate there is a prime

lfloor(δ minus δprime)N2rfloor le N prime le 2lfloor(δ minus δprime)N2rfloor

Partitioning 1⋯N primelfloorNN primerfloor into lfloorNN primerfloor chunks Cj of size N prime we find that forleast one of the chunks ∣A cap Cj ∣ ge δprimeN prime and hence A cap Cj contains a three-termarithmetic progression

Third to make use of the Fourier transform we shall work modulo N Torule out ldquowraparoundrdquo (ie to make sure that the three-term progression we willhave found is not say N minus 4N minus 12) we consider several cases If

∣a isin A ∶ a le N3∣ gt δ(1 + δ20)∣A∣3 (22)

the assumption Rothδ(1+δ25) ensures that there is an arithmetic progression inA1 = a isin A ∶le N1 = lfloorN3rfloor We can argue similarly if

∣a isin A ∶ a gt 2N3∣ gt δ(1 + δ20)∣A∣3 (23)

Therefore we assume for the rest of the proof that (22) and (23) fail ie that

B = a isin A ∶ N3 lt a le 2N3

is of cardinality ∣B∣ ge δ(1minus δ10)∣A∣3 Observe that if a isin A and b c isin B are suchthat a + c equiv 2b mod N then either (a b c) is a proper arithmetic progression(without wraparound) or a = b = c

Now the Fourier-analytic part begins The triples (a b c) with a + c equiv 2bmod N can be counted as follows

S = sumaisinAbcisinB

1a+c=2b =N

sumabc=1

1A(a)1B(b)1B(c)1a+c=2b (24)

= 1

N

N

sumabc=1

Nminus1

sump=0

1A(a)1B(b)1B(c) exp(minus2πip(a minus 2b + c)N) (25)

= 1

N

Nminus1

sump=0

1A(p)1B(minus2p)1B(p) (26)

Observing that 1A(0) = ∣A∣ and similarly for BC and using the CauchyndashSchwarz

5

inequality and that N is odd we get

S ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(minus2p)1B(p)∣ (27)

ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(p)∣2 (28)

= 1

N∣A∣ ∣B∣2 minusmax

pne0∣1A(p)∣ ∣B∣ (29)

This may be understood as follows Among the sim N arithmetic progressions(a b c) sim ∣A∣N have a isin A and similarly sim ∣B∣N have b isin B or c isin B If theseevents would be independent (which is the case if A is random) we would haveS sim N 3 times (∣A∣N) times (∣B∣N)2 = ∣A∣∣B∣2N If the non-zero Fourier coefficients of1A are small A has no distinguished patterns and therefore we expect this tobe a good approximation

Proceeding with the argument consider two cases If maxpne0 ∣1A(p)∣ le δ∣A∣6then

(29) ge ∣A∣∣B∣(∣B∣N minus δ6) ge ∣A∣∣B∣(δ(1 minus δ10)3 minus δ6) ge c(δ)N 2

which is much greater than N and we are doneNow assume that there exists p ne 0 such that ∣1A(p)∣ gt δ∣A∣6 (Recall that this

means that A is correlated with some pattern of period p) In this case we shalldecompose 1⋯N into several arithmetic progressions so that the density ofA in at least one of them is at least δ(1 + δ50) Formally let m = lceil100

δ rceil and let

Pj = 0 le x le N minus 1 ∶ jNm

le px lt (j + 1)Nm

0 le j ltm

These are arithmetic progressions modulo N indeed if r be such that pr equiv 1mod N then

Pj = kr ∶ jNm

le k lt (j + 1)Nm

Then

∣1A(p)∣ = ∣sumjsumxisinPj

1A(x) exp(minus2πipxN)∣ (210)

le ∣sumjsumxisinPj

1A(x) exp(minus2πijm)∣ + 2π

msumjsumxisinPj

1A(x) (211)

=RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRR+ 2π∣A∣

m (212)

6

Hence RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRRge ∣1A(p)∣ minus

2π∣A∣m

ge δ∣A∣12

(213)

If all the sets AcapPj were of the same size the sum would vanish Quantitativelywe have

Exercise 29 Prove if f ∶ ZmZrarr R and f = 1msum

mminus1j=0 f(j) then

mminus1maxj=0

f(j) ge f + 1

2m∣mminus1

sumj=0

f(j) exp(minus2πijm)∣ (214)

Applying (214) to f(j) = ∣A cap Pj ∣ we obtain that there exists j such that

∣A cap Pj ∣ ge∣A∣m

+ δ∣A∣24m

ge δ(1 + δ25)∣Pj ∣

Now we further decompose this Pj into proper arithmetic progressions Among

the lfloorradicNrfloor numbers r2r⋯ lfloor

radicNrfloorr there are two which are not far from one

another rlowast = k2r minus k1r mod N isin 1⋯ lfloorradicNrfloor Let klowast = k2 minus k1 note that

klowast leradicN For each 1 le k le klowast split Pj into klowast arithmetic progressions of step

rlowast (still possibly with wraparound) and then split each of these into proper

arithmetic progressions without wraparound All but at most CradicN of these

progressions have length ge cradicN hence at least one of them P prime

j sub Pj satisfies

∣P primej ∣ ge c

radicN ∣AcapP prime

j ∣ ge δ(1+δ25)∣P primej ∣ By the induction assumption AcapP prime

j containsa three-term arithmetic progression

Exercise 210 Show that

1 N0(δ) le exp exp(Cδ)

2 the assumption ∣A∣ ge δN can be replaced with ∣A∣ ge C1N log logN for asufficiently large constant C1

23 Fast Fourier transform

The naıve implementation of the discrete Fourier transform on ZNZ requiressim constN 2 arithmetic operations Fast Fourier transform allows to do the jobwith O(N logN) operations See Montgomery [2014] for a historical discussion

Denote the smallest number of arithmetic operations required for the compu-tation of the Fourier transform by n(N)

Proposition 211 n(2N) le 2n(N) + 3N

7

Proof Let f ∶ Z2NZrarr C Construct two auxiliary functions g h ∶ ZNZrarr C

g(n) = f(2n) h(n) = f(2n + 1)

Then

f(p) =2Nminus1

sumn=0

f(n) exp(minus2πinp(2N))

=Nminus1

sumn=0

f(2n) exp(minus2πi2np(2N)) +Nminus1

sumn=0

f(2n + 1) exp(minus2πi (2n + 1)p(2N))

= g(p) + exp(minusπipN)h(p)

which means that for 0 le p le N minus 1

f(p) = g(p)+ exp(minusπipN)h(p) f(p+N) = g(p)minus exp(minusπipN)h(p) (215)

The computation of f from g and h using (215) requires 3N arithmetic opera-tions (if we assume that the values of exp(minusπipN) can be precomputed)

Corollary 212 n(2n) le (32)n2n for n ge 1

Proof By induction

This corollary implies that n(N) le CN logN when N is a power of 2 Similarreasoning applies to N which do not have large prime factors The general caserequires a separate argument for large prime N which we do not reproduce here

Applications From item 2 of Exercise 27 we obtain that the convolution f lowastg =(f g)or of two functions f g ∶ ZNZ rarr C can be computed in O(n(N)) steps forN = 2n this is O(N logN)

Next p(x) = sumdk=0 akxk and q(x) = sumdk=0 bkx

k be two polynomials Their productis equal to (pq)(x) = sum2d

k=0(sumkl=0 albkminusl)xk This looks quite similar to a convolu-tion except that there is no wraparound

Exercise 213 The product of two polynomials of degree d can be computed inO(d log d) arithmetic operations

Exercise 214 Long multiplication of two d-digit numbers takes O(d2) operations(where an operation means addition or multiplication of single-digit numbers)Propose an algorithm which would only require O(d log10 d) operations

8

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

inequality and that N is odd we get

S ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(minus2p)1B(p)∣ (27)

ge 1

N∣A∣ ∣B∣2 minusmax

pne0

1

N∣1A(p)∣ times

Nminus1

sump=1

∣1B(p)∣2 (28)

= 1

N∣A∣ ∣B∣2 minusmax

pne0∣1A(p)∣ ∣B∣ (29)

This may be understood as follows Among the sim N arithmetic progressions(a b c) sim ∣A∣N have a isin A and similarly sim ∣B∣N have b isin B or c isin B If theseevents would be independent (which is the case if A is random) we would haveS sim N 3 times (∣A∣N) times (∣B∣N)2 = ∣A∣∣B∣2N If the non-zero Fourier coefficients of1A are small A has no distinguished patterns and therefore we expect this tobe a good approximation

Proceeding with the argument consider two cases If maxpne0 ∣1A(p)∣ le δ∣A∣6then

(29) ge ∣A∣∣B∣(∣B∣N minus δ6) ge ∣A∣∣B∣(δ(1 minus δ10)3 minus δ6) ge c(δ)N 2

which is much greater than N and we are doneNow assume that there exists p ne 0 such that ∣1A(p)∣ gt δ∣A∣6 (Recall that this

means that A is correlated with some pattern of period p) In this case we shalldecompose 1⋯N into several arithmetic progressions so that the density ofA in at least one of them is at least δ(1 + δ50) Formally let m = lceil100

δ rceil and let

Pj = 0 le x le N minus 1 ∶ jNm

le px lt (j + 1)Nm

0 le j ltm

These are arithmetic progressions modulo N indeed if r be such that pr equiv 1mod N then

Pj = kr ∶ jNm

le k lt (j + 1)Nm

Then

∣1A(p)∣ = ∣sumjsumxisinPj

1A(x) exp(minus2πipxN)∣ (210)

le ∣sumjsumxisinPj

1A(x) exp(minus2πijm)∣ + 2π

msumjsumxisinPj

1A(x) (211)

=RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRR+ 2π∣A∣

m (212)

6

Hence RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRRge ∣1A(p)∣ minus

2π∣A∣m

ge δ∣A∣12

(213)

If all the sets AcapPj were of the same size the sum would vanish Quantitativelywe have

Exercise 29 Prove if f ∶ ZmZrarr R and f = 1msum

mminus1j=0 f(j) then

mminus1maxj=0

f(j) ge f + 1

2m∣mminus1

sumj=0

f(j) exp(minus2πijm)∣ (214)

Applying (214) to f(j) = ∣A cap Pj ∣ we obtain that there exists j such that

∣A cap Pj ∣ ge∣A∣m

+ δ∣A∣24m

ge δ(1 + δ25)∣Pj ∣

Now we further decompose this Pj into proper arithmetic progressions Among

the lfloorradicNrfloor numbers r2r⋯ lfloor

radicNrfloorr there are two which are not far from one

another rlowast = k2r minus k1r mod N isin 1⋯ lfloorradicNrfloor Let klowast = k2 minus k1 note that

klowast leradicN For each 1 le k le klowast split Pj into klowast arithmetic progressions of step

rlowast (still possibly with wraparound) and then split each of these into proper

arithmetic progressions without wraparound All but at most CradicN of these

progressions have length ge cradicN hence at least one of them P prime

j sub Pj satisfies

∣P primej ∣ ge c

radicN ∣AcapP prime

j ∣ ge δ(1+δ25)∣P primej ∣ By the induction assumption AcapP prime

j containsa three-term arithmetic progression

Exercise 210 Show that

1 N0(δ) le exp exp(Cδ)

2 the assumption ∣A∣ ge δN can be replaced with ∣A∣ ge C1N log logN for asufficiently large constant C1

23 Fast Fourier transform

The naıve implementation of the discrete Fourier transform on ZNZ requiressim constN 2 arithmetic operations Fast Fourier transform allows to do the jobwith O(N logN) operations See Montgomery [2014] for a historical discussion

Denote the smallest number of arithmetic operations required for the compu-tation of the Fourier transform by n(N)

Proposition 211 n(2N) le 2n(N) + 3N

7

Proof Let f ∶ Z2NZrarr C Construct two auxiliary functions g h ∶ ZNZrarr C

g(n) = f(2n) h(n) = f(2n + 1)

Then

f(p) =2Nminus1

sumn=0

f(n) exp(minus2πinp(2N))

=Nminus1

sumn=0

f(2n) exp(minus2πi2np(2N)) +Nminus1

sumn=0

f(2n + 1) exp(minus2πi (2n + 1)p(2N))

= g(p) + exp(minusπipN)h(p)

which means that for 0 le p le N minus 1

f(p) = g(p)+ exp(minusπipN)h(p) f(p+N) = g(p)minus exp(minusπipN)h(p) (215)

The computation of f from g and h using (215) requires 3N arithmetic opera-tions (if we assume that the values of exp(minusπipN) can be precomputed)

Corollary 212 n(2n) le (32)n2n for n ge 1

Proof By induction

This corollary implies that n(N) le CN logN when N is a power of 2 Similarreasoning applies to N which do not have large prime factors The general caserequires a separate argument for large prime N which we do not reproduce here

Applications From item 2 of Exercise 27 we obtain that the convolution f lowastg =(f g)or of two functions f g ∶ ZNZ rarr C can be computed in O(n(N)) steps forN = 2n this is O(N logN)

Next p(x) = sumdk=0 akxk and q(x) = sumdk=0 bkx

k be two polynomials Their productis equal to (pq)(x) = sum2d

k=0(sumkl=0 albkminusl)xk This looks quite similar to a convolu-tion except that there is no wraparound

Exercise 213 The product of two polynomials of degree d can be computed inO(d log d) arithmetic operations

Exercise 214 Long multiplication of two d-digit numbers takes O(d2) operations(where an operation means addition or multiplication of single-digit numbers)Propose an algorithm which would only require O(d log10 d) operations

8

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Hence RRRRRRRRRRRsumj

∣A cap Pj ∣ exp(minus2πijm)RRRRRRRRRRRge ∣1A(p)∣ minus

2π∣A∣m

ge δ∣A∣12

(213)

If all the sets AcapPj were of the same size the sum would vanish Quantitativelywe have

Exercise 29 Prove if f ∶ ZmZrarr R and f = 1msum

mminus1j=0 f(j) then

mminus1maxj=0

f(j) ge f + 1

2m∣mminus1

sumj=0

f(j) exp(minus2πijm)∣ (214)

Applying (214) to f(j) = ∣A cap Pj ∣ we obtain that there exists j such that

∣A cap Pj ∣ ge∣A∣m

+ δ∣A∣24m

ge δ(1 + δ25)∣Pj ∣

Now we further decompose this Pj into proper arithmetic progressions Among

the lfloorradicNrfloor numbers r2r⋯ lfloor

radicNrfloorr there are two which are not far from one

another rlowast = k2r minus k1r mod N isin 1⋯ lfloorradicNrfloor Let klowast = k2 minus k1 note that

klowast leradicN For each 1 le k le klowast split Pj into klowast arithmetic progressions of step

rlowast (still possibly with wraparound) and then split each of these into proper

arithmetic progressions without wraparound All but at most CradicN of these

progressions have length ge cradicN hence at least one of them P prime

j sub Pj satisfies

∣P primej ∣ ge c

radicN ∣AcapP prime

j ∣ ge δ(1+δ25)∣P primej ∣ By the induction assumption AcapP prime

j containsa three-term arithmetic progression

Exercise 210 Show that

1 N0(δ) le exp exp(Cδ)

2 the assumption ∣A∣ ge δN can be replaced with ∣A∣ ge C1N log logN for asufficiently large constant C1

23 Fast Fourier transform

The naıve implementation of the discrete Fourier transform on ZNZ requiressim constN 2 arithmetic operations Fast Fourier transform allows to do the jobwith O(N logN) operations See Montgomery [2014] for a historical discussion

Denote the smallest number of arithmetic operations required for the compu-tation of the Fourier transform by n(N)

Proposition 211 n(2N) le 2n(N) + 3N

7

Proof Let f ∶ Z2NZrarr C Construct two auxiliary functions g h ∶ ZNZrarr C

g(n) = f(2n) h(n) = f(2n + 1)

Then

f(p) =2Nminus1

sumn=0

f(n) exp(minus2πinp(2N))

=Nminus1

sumn=0

f(2n) exp(minus2πi2np(2N)) +Nminus1

sumn=0

f(2n + 1) exp(minus2πi (2n + 1)p(2N))

= g(p) + exp(minusπipN)h(p)

which means that for 0 le p le N minus 1

f(p) = g(p)+ exp(minusπipN)h(p) f(p+N) = g(p)minus exp(minusπipN)h(p) (215)

The computation of f from g and h using (215) requires 3N arithmetic opera-tions (if we assume that the values of exp(minusπipN) can be precomputed)

Corollary 212 n(2n) le (32)n2n for n ge 1

Proof By induction

This corollary implies that n(N) le CN logN when N is a power of 2 Similarreasoning applies to N which do not have large prime factors The general caserequires a separate argument for large prime N which we do not reproduce here

Applications From item 2 of Exercise 27 we obtain that the convolution f lowastg =(f g)or of two functions f g ∶ ZNZ rarr C can be computed in O(n(N)) steps forN = 2n this is O(N logN)

Next p(x) = sumdk=0 akxk and q(x) = sumdk=0 bkx

k be two polynomials Their productis equal to (pq)(x) = sum2d

k=0(sumkl=0 albkminusl)xk This looks quite similar to a convolu-tion except that there is no wraparound

Exercise 213 The product of two polynomials of degree d can be computed inO(d log d) arithmetic operations

Exercise 214 Long multiplication of two d-digit numbers takes O(d2) operations(where an operation means addition or multiplication of single-digit numbers)Propose an algorithm which would only require O(d log10 d) operations

8

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Proof Let f ∶ Z2NZrarr C Construct two auxiliary functions g h ∶ ZNZrarr C

g(n) = f(2n) h(n) = f(2n + 1)

Then

f(p) =2Nminus1

sumn=0

f(n) exp(minus2πinp(2N))

=Nminus1

sumn=0

f(2n) exp(minus2πi2np(2N)) +Nminus1

sumn=0

f(2n + 1) exp(minus2πi (2n + 1)p(2N))

= g(p) + exp(minusπipN)h(p)

which means that for 0 le p le N minus 1

f(p) = g(p)+ exp(minusπipN)h(p) f(p+N) = g(p)minus exp(minusπipN)h(p) (215)

The computation of f from g and h using (215) requires 3N arithmetic opera-tions (if we assume that the values of exp(minusπipN) can be precomputed)

Corollary 212 n(2n) le (32)n2n for n ge 1

Proof By induction

This corollary implies that n(N) le CN logN when N is a power of 2 Similarreasoning applies to N which do not have large prime factors The general caserequires a separate argument for large prime N which we do not reproduce here

Applications From item 2 of Exercise 27 we obtain that the convolution f lowastg =(f g)or of two functions f g ∶ ZNZ rarr C can be computed in O(n(N)) steps forN = 2n this is O(N logN)

Next p(x) = sumdk=0 akxk and q(x) = sumdk=0 bkx

k be two polynomials Their productis equal to (pq)(x) = sum2d

k=0(sumkl=0 albkminusl)xk This looks quite similar to a convolu-tion except that there is no wraparound

Exercise 213 The product of two polynomials of degree d can be computed inO(d log d) arithmetic operations

Exercise 214 Long multiplication of two d-digit numbers takes O(d2) operations(where an operation means addition or multiplication of single-digit numbers)Propose an algorithm which would only require O(d log10 d) operations

8

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

3 Fourier series

31 Fourier coefficients and Plancherel theorem

From the finite group ZNZ we pass to the torus T = RZ (defined in the naturalway eg [13] + [34] = [112]) We identify functions on the torus with 1-periodic functions of a real variable

A character of T is a continuous function χ ∶ Trarr Ctimes such that

χ(x + y) = χ(x)χ(y) (31)

Exercise 31 Any character of T is of the form ep(x) = exp(2πipx) for some p isin Z

The goal is to expand f in a series in ep The questions of convergence are del-icate as we shall see We start with the discussion of mean-square convergencefor which the theory is simplest We think of ⟨f g⟩ = int

10 f(x)g(x)dx as an inner

product however we should keep in mind that if we work with Riemann inte-grals the space of square-integrable functions is not complete We shall focus onpiecewise-continuous functions ie [01] is a finite union of closed intervals suchthat the function is uniformly continuous in the interior of each of them andtry to extract the most of the pre-Hilbert point of view (Ultimately a betterapproach would be to work with Lebesgue integrals and enjoy the benefits of aproper Hilbert space)

The following exercise shows that ep are orthonormal

Exercise 32 int1

0 ep(x)eq(x)dx =⎧⎪⎪⎨⎪⎪⎩

1 p = q0

This leads us to

Definition 33 Let f ∶ T rarr C be piecewise continuous The numbers f(p) =⟨f ep⟩ are called the Fourier coefficients of f and the formal2 series

infin

sump=minusinfin

f(p)epis called the Fourier series representing f

The discussion of convergence starts with

Theorem 34 (Plancherel) Let f ∶ Trarr C be piecewise continuous Then

infin

sump=minusinfin

∣f(p)∣2 = int1

0∣f(x)∣2dx

Proof One direction holds for any orthonormal system

2here formal refers to manipulations with formulaelig disregarding questions of convergence

9

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Claim 35 Let H be an inner product space and let (uα)αisinA be an orthonormalsystem of vectors Then for any f isinH

sumαisinA

∣⟨f uα⟩∣2 le ∥f∥2 = ⟨f f⟩

Proof of Claim Observe that for any finite Aprime sub A

0 le ∥f minus sumαisinAprime

⟨f uα⟩uα∥2 = ∥f∥2 minussumαisinA

∣⟨f uα⟩∣2 le ∥f∥2

The inequality le of the theorem follows by applying the claim to the space ofpiecewise continuous functions The other direction ge requires rolling up thesleeves Let f be piecewise continuous and let M = max ∣f ∣ Fix ε gt 0 andchoose f1 isin C(T) such that max ∣f1∣ =M and f = f1 outside a union of intervalsof total length le ε Now choose3 a trigonometric polynomial f2 = sumNp=minusN cpep suchthat max ∣f1 minus f2∣ le ε Then (still using the notation ∥f∥2 = ⟨f f⟩)

∥f∥ le ∥f2∥ + ∥f1 minus f2∥ + ∥f minus f1∥ le ∥f2∥ + ε + 2Mradicε

whereas (similarly)

iquestAacuteAacuteAgrave

N

sump=minusN

∣f(p)∣2 ge

iquestAacuteAacuteAgrave

N

sump=minusN

∣f2(p)∣2 minus ε minus 2Mradicε = ∥f2∥ minus ε minus 2M

radicε

ie iquestAacuteAacuteAgrave

infin

sump=minusinfin

∣f(p)∣2 ge ∥f∥ minus 2(ε + 2Mradicε)

for any ε gt 0

The Plancherel theorem implies mean-square convergence of Fourier series

Exercise 36 If f ∶ Trarr C is piecewise continuous then

limNrarrinfin

int1

0∣f(x) minus

N

sump=minusN

f(p)ep(x)∣2dx = 0 (32)

We remind that (32) does not imply pointwise convergence and in fact doesnot even imply the existence of one point at which the series converges We shallreturn to the question of pointwise convergence later

Here is another corollary of the Plancherel theorem

Corollary 37 If f g ∶ T rarr C are piecewise continuous and f equiv g then f(x) =g(x) at all the continuity points of f 3We shall prove the Weierstrass approximation theorem in Section 41 without circular reasoning

10

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Exercise 38 If f g ∶ Trarr C are piecewise continuous then

infin

sump=minusinfin

f(p)g(p) = ⟨f g⟩

Let us see a couple of applications before we move on More applications ofFourier series will be discussed in the next section

The value of ζ(2) Let f(x) = x (for x isin (01)) so that ∥f∥2 = 13 The Fourier

coefficients of f can be computed as follows

f(p) = int1

0x exp(minus2πipx)dx = minus 1

2πip int1

0xd

dxexp(minus2πipx)dx

= minus 1

2πipx exp(minus2πipx)∣1

0+ 1

2πip int1

0exp(minus2πipx)dx = minus 1

2πip

(33)

for p ne 0 and f(0) = 12 Hence

1

4+ 2

infin

sump=1

1

4π2p2= 1

3 ζ(2) =

infin

sump=1

1

p2= π

2

6

Exercise 39 Compute ζ(4) = suminfinp=1 p

minus4

Poincare inequality Here is another application

Exercise 310 Prove that for any reasonable (differentiable with piecewise con-tinuous derivative) f ∶ Trarr C

int1

0∣f(x) minus f ∣2dx le 1

4π2 int1

0∣f prime(x)∣2dx where f = int

1

0f(x)dx (34)

The right-hand side of (34) measures the global magnitude of fluctuations (itis the variance of f(X) where X isin T is chosen uniformly at random) whereasthe right-hand side measures the local fluctuations To appreciate the mean-ing of such estimates it may be useful to consider the following d-dimensionalgeneralisation of (34) for any reasonable f ∶ Td rarr C

intTd

∣f(x) minus f ∣2dx le 1

4π2 intTd∥nablaf(x)∥2dx where f = int

Tdf(x)dx (35)

(and ∥nablaf∥2 = sumdj=1 ∣partfpartxj

∣2) In particular if f satisfies the Lipschitz estimate

∣f(x) minus f(y)∣ le dist(x y)

11

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

where dist is the usual Euclidean distance on the d-dimensional torus then thevariance of f(X) (where X isin Td is chosen uniformly at random) is bounded by1(4π2) This estimate is much better than the naıve bound

intTd

∣f(x) minus f ∣2dx le diam2 Td = d4

which grows with the dimension (it is instructive to compare the two bounds forthe simple special case f(x) = sumdj=1 fj(xj)) This is an instance of the concentra-tion phenomenon in high dimension put forth in the 1970s by V Milman SeeGiannopoulos and Milman [2000] Ledoux [2001] and references therein

32 Convergence of Fourier series

We return to the main narrative Let f ∶ Trarr C be piecewise continuous We areinterested in the convergence of partial sums

N

sump=minusN

f(p)ep(x) (36)

to f(x) We first evaluate

(36) =N

sump=minusNint f(y)ep(y)dy ep(x)

= int f(y)⎛⎝

N

sump=minusN

ep(x minus y)⎞⎠dy = (DN lowast f)(y)

where

DN(x) =N

sump=minusN

ep(x) =sin((2N + 1)πx)

sin(πx)(37)

is the Dirichlet kernel (see Figure 41)

Lemma 311 (RiemannndashLebesgue) Let f ∶ T rarr C be piecewise continuous

Then f(p)rarr 0 as prarr plusmninfin

The intuitive explanation may go as follows If a function is constant f(p) = 0for all p ne 0 A typical function eg f(x) = dist(xZ) is not constant howeverfor large p it is almost constant on scale 1p which means that the p-th Fouriercoefficient should be small The following argument formalises this idea

12

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

minus04 minus02 0 02 04minus20

0

20

40

60

80

x

DN(x

)

D10(x)D50(x)

Figure 31 The Dirichlet kernel for N = 1050

Proof We perform a change of variables xrarr x + 12p

f(p) = int f(x)ep(x)dx

= int f(x + 1

2p)ep(x +

1

2p)dx

= int f(x)ep(x +1

2p)dx + int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx

The first term on the right-hand side is equal to the negative of the left-handside The second term is bounded by

∣int (f(x) minus f(x + 1

2p))ep(x +

1

2p)dx∣ le int ∣f(x) minus f(x + 1

2p)∣dx

therefore

∣f(p)∣ le 1

2 int∣f(x) minus f(x + 1

2p)∣dx

and this tends to zero as prarr plusmninfin (why)

Now we can prove

Proposition 312 Let f ∶ Trarr C be piecewise continuous and suppose x0 isin T issuch that there exist a isin (01] and C gt 0 for which

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a (38)

Then (DN lowast f)(x0)rarr f(x0) as N rarrinfin

13

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Proof Without loss of generality f is real-valued and x0 = 0 f(x0) = 0 Decom-pose

int f(x)DN(x)dx = int∣x∣leδ

+intδlexle 1

2

Recalling that ∣ sin(πx)∣ ge 2∣x∣ for ∣x∣ le 12 the first integral is bounded by

∣int∣x∣leδ

∣ le int∣x∣leδ

C ∣x∣a dx2∣x∣

le C primeδa

The second integral is equal to

intδlexle 1

2

= int12

minus 12

[f1(x) sin(2Nπx) + f2(x) cos(2Nπx)]dx (39)

where f1(x) = f(x) cot(πx)1∣x∣geδ and f2(x) = f(x)1∣x∣geδ These two functionsare piecewise continuous (for any δ gt 0) hence the RiemannndashLebesgue lemmaimplies that (39)rarr 0 as N rarrinfin It remains to let δ rarr +0

(Pisa Italy)

Exercise 313 Amplify Proposition 312 as follows iff ∶ Trarr C is piecewise continuous and there exists A isin Csuch that

int12

0

∣f(x0 + t) + f(x0 minus t) minus 2A∣∣t∣

dt ltinfin (310)

then (DN lowast f)(x0) rarr A as N rarr infin If f satisfies (38)then (310) holds with A = f(x0) What happens if fhas a jump continuity at x0 eg f = 1[012] What isthe connection to the picture on the left

Another important corollary of the RiemannndashLebesgue lemma is the Riemannlocalisation pronciple which asserts that the convergence of the Fourier seriesat a point only depends on the behaviour of the function in the vicinity of thispoint

Exercise 314 If f g ∶ Trarr C are piecewise continuous functions that coincide ina neighbourhood of x0 then (DN lowast f)(x0)rarr A iff (DN lowast g)(x0)rarr A

Remark 315 The RiemannndashLebesgue lemma does not provide any quantitativerate of decay This can not be improved without introducing extra assumptions

Exercise 316 Let (ε(p))pge1 be a sequence of positive numbers tending to zeroas prarrinfin Show that one can choose 1 le p1 lt p2 lt ⋯ such that

f(x) =infin

sump=1

ε(pj)epj(x)

is a continuous function Clearly ε(p)minus1∣f(p)∣ ETHrarr 0 as prarrinfin

14

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Divergence If f is only assumed to be continuous the Fourier series may di-verge That is f is not the pointwise sum of its Fourier series

Proposition 317 (Du Bois-Reymond) There exists f isin C(T) such that (DN lowastf)(0) diverges as N rarrinfin

The proof is based on the estimate

int ∣DN(x)∣dx = int12

minus 12

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

gt 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣∣ sinπx∣

dx

ge 2N

sumk=1int

k2N+1

kminus12N+1

∣ sin((2N + 1)πx)∣πk(2N + 1)

dx ge c logN rarrinfin

(311)

In particular one can find a continuous function fN with sup ∣fN ∣ le 1 and

int DN(x)fN(x)dxrarrinfin as N rarrinfin4

The idea is to construct fN which will approximate sign(DN) along a subse-quence of N and such that the Fourier coefficients of fN are small when p is farfrom N and then make a convergent series of of fNk (when Nk are sparse enoughto avoid interference between different FNk) This is implemented as follows

Proof LetfN(x) = sin((2N + 1)πx) 0 le x lt 1

and continue it periodically

Exercise 318 Prove that int DN(x)fN(x)dx ge c1 logN

For M not very close to M int DN(x)fM(x)dx is not very large as madequantitative by

Exercise 319 ∣ int DN(x)fM(x)dx∣ le⎧⎪⎪⎨⎪⎪⎩

CMN M le N2CNM M ge 2N

Now set

f(x) =infin

sumk=1

1

k2fNk(x)

where we shall take Nk = 3k3 According to Exercises 318 and 319

(DNk lowast f)(0) gec1 logNk

k2minuskminus1

sumj=1

1

j2

CNj

Nkminus

infin

sumj=k+1

1

j2

CNk

Njge c2k

4One can prove the theorem using this but property of DN if one relies on the BanachndashSteinhaus theorem fromfunctional analysis The advantage of the more classical proof reproduced below is that it gives an explicitestimate on the rate of divergence

15

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

33 Fejer summation

The negative result of the previous paragraph is somewhat surprising giventhat by Weierstrass theorem any continuous function on T can be uniformlyapproximated by trigonometric polynomials Of course DN lowastf is the orthogonalprojection of f onto the space of trigonometric polynomials of degree le N ie it isthe trigonometric polynomial PN of degree N which minimises int ∣PNminusf ∣2dx Thismeans that the best approximation in the norm ∥sdot∥ = ∥sdot∥2 may not be a very goodapproximations in the maximum norm However good uniform approximationscan be constructed from DN lowast f using the following simple procedure

Theorem 320 (Fejer) If f isin C(T) then

1

N

Nminus1

sumn=0

(Dn lowast f) =Nminus1

sump=minusN+1

(1 minus ∣p∣N)f(p)ep f as N rarrinfin

The arithmetic mean appearing in the left-hand side is known as Cesaro sum-mation If (an)nge0 is a sequence of numbers one constructs the Cesaro meansAN = 1

N sumNminus1n=0 an

Exercise 321 If (an) converges to a limit L then also AN rarr L

This procedure is useful since there are divergent sequences such as an = (minus1)nfor which the Cesaro means converge the limit may be considered as a generalisedlimit of (an) It is remarkable that such a simple resummation procedure workssimultaneously for all continuous functions

Proof of Theorem 320 We first observe that 1N sum

Nn=0(Dn lowast f) = SN lowast f where

SN(x) = 1

N

Nminus1

sumn=0

Dn(x) =1

N

Nminus1

sumn=0

sin((2N + 1)πx)sin(πx)

= 1

N

sin2(Nπx)sin2(πx)

The Fejer kernel SN boasts the following properties

1 SN ge 0

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12SN(x)dxrarr 0 as N rarrinfin

From these properties

∣f(x) minus (SN lowast f)(x)∣ =2∣int (f(x) minus f(x minus y))SN(y)dy∣

le1int ∣f(x) minus f(x minus y)∣SN(y)dy

16

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Fix ε gt 0 and choose δ so that ∣f(x) minus f(xprime)∣ le ε2 for ∣x minus xprime∣ le δ and then use 3

to choose N0 so that for N ge N0

intδle∣y∣le 1

2

SN(y)dy le ε

2 max ∣f ∣

Then for N ge N0

int ∣f(x) minus f(x minus y)∣SN(y)dy = int∣y∣leδ

+intδle∣y∣le 1

2

le ε2+ ε

2= ε

Remark 322 A kernel SN(x) satisfying 1ndash3 5 is called a non-negative summa-bility kernel A kernel is called a (general) summability kernel if

1 supN int ∣SN(x)∣dx ltinfin

2 int1

0 SN(x)dx = 1

3 For any δ gt 0 intδle∣x∣le 12∣SN(x)∣dxrarr 0 as N rarrinfin

Note that any non-negative summability kernel is a summability kernel

Exercise 323 Show that the conclusion SN lowast f f of Fejerrsquos theorem holds forany summability kernel SN

Here is one additional example of a summability kernel We shall see anotherone in Section 41 and yet another one in Section 42

Exercise 324 Prove the following

1 The Poisson kernel

Pr(x) =R [1 + r exp(2πix)1 minus r exp(2πix)

] = 1 minus r2

1 minus 2r cos(2πx) + r2 r isin [01)

satisfies that (P1minusη)ηrarr+0 is a non-negative summability kernel

2 If f isin C(T ) thenu(re2πix) = (Pr lowast f)(x)

is continuous in the closed unit disk in R2 and satisfies ∆u = 0 in the interiorof the disk (ie it is harmonic)

5or more precisely a family of kernels depending on an additional large parameter N rarrinfin To emphasise thiswe may sometimes write (SN)Nrarrinfin We shall also consider kernels (Sη)ηrarr+0 depending on a small parameterη rarr +0 the definitions are then adjusted in an obvious way

17

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

4 Fourier series further topics and applications

41 Polynomial approximation

As we noted Fejerrsquos theorem implies the Weierstrass approximation theoremfor any f isin C(T)

limNrarrinfin

EN(f) = 0 where EN(f) = infcminusN ⋯cNisinC

∥f minusN

sump=minusN

cpep∥infin (41)

Moreover similar arguments yield good quantitative bounds The followingpropositions are based on the work of Dunham Jackson and are known as Jack-sonrsquos theorems

Theorem 41 For any f isin C(T) EN(f) le Cωf( 1N ) where

ωf(δ) = sup∣xminusxprime∣leδ

∣f(x) minus f(xprime)∣ (42)

is the modulus of continuity

Proof Let JN(x) be a non-negative trigonometric polynomial of degree say 2Nsuch that int JN(x)dx = 1 Then

∣(JN lowast f)(x) minus f(x)∣ le int12

minus 12

∣f(x) minus f(x minus y)∣JN(y)dy

=N

sumj=1int jminus1

2Nle∣y∣le j

2N

∣f(x) minus f(x minus y)∣JN(y)dy

The j-th integral is bounded by

ωf(j(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy le jωf(1(2N))int jminus12N

le∣y∣le j2N

∆N(y)dy

hence

E2N(f) le ωf(1(2N))N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy

If we find JN such that

supN

N

sumj=1

j int jminus12N

le∣y∣le j2N

JN(y)dy ltinfin (43)

the sum to be bounded uniformly N This condition is strictly stronger thanproperty 3 of non-negative summability kernels (43) implies 3 (why) but

18

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

not vice versa since it does not hold for the Fejer kernel SN (why) Thereforewe take the Jackson kernel

JN(x) = 3

2N 3 +Nsin4(πNx)sin4(πx)

= 3N

2N 2 + 1SN(x)2

The faster decay ensures that (43) holds (why) The second representationensures that indeed this is a trigonometric polynomial of degree 2N The nor-malisation constant is chosen to ensure that the integral is one

int12

minus 12

SN(x)2dx =Nminus1

sump=minusN+1

(1 minus ∣p∣N)2

= 1 + 2

N 2

Nminus1

sump=1

p2

= 1 + (N minus 1)(2N minus 1)3N

= 1 + 2N 2

3N

Thus E2N(f) le Cωf(1(2N))

Exercise 42 Prove if f isin Ck(T) for some k ge 1 then EN(f) le CkNminuskωf (k)( 1N )

Exercise 43 Let f isin C(T) Then

sup ∣f minusDN lowast f ∣ le C logN ωf(1N)

In particular we recover the DinindashLipschitz criterion DN lowast f f wheneverωf(t) log 1

t rarr 0 as trarr +0

Exercise 44 State and prove a version of Theorem 41 for the approximation ofa function in C[minus11] by algebraic polynomials of degree le N You are welcometo rely on Theorem 41 if needed

Constructive function theory In the early XXth century Serge Bernstein putforth the idea that the analytic properties of a function such as continuity andsmoothness are related to the rate of approximation by polynomials (or trigono-metric polynomials) This circle of questions and results known as constructivefunction theory For example f ∶ T rarr C is continuous if and only if EN(f) rarr 0As a more sophisticated illustration we prove the following converse to Jacksonrsquostheorem (Theorem 41)

Theorem 45 (Bernstein) Let a isin (01) For any f ∶ Trarr C

supδisin(0 1

2]

ωf(δ)δminusa le Ca supNge1

EN(f)Na Ca = 2π21minusa + 1

21minusa minus 1 (44)

where EN are as in (41) In particular the left-hand side is finite whenever theright-hand side is finite

19

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Remark 46 Note that we assume a lt 1 For a = 1 the statement does not holdand in fact EN(f) = O(1N) iff

supxisinT hisin(0 1

2)

∣f(x + h) + f(x minus h) minus 2f(x)∣h

ltinfin

The proof relies on the following very useful inequality also due to Bernstein

Theorem 47 (Bernstein) Let P (x) = sumNp=minusN cpep(x) be a trigonometric polyno-mial of degree N Then ∥P prime∥infin le 2πN∥P ∥infin

Proof of Theorem 47 (cf Montgomery [2014]) We look for 2N coefficients a1⋯ a2N

(independent of P but dependent on N) such that

P prime(0) =2N

sumk=1

akP (2k minus 1

4N) (45)

so that we can bound ∣P prime(0)∣ le sum2Nk=1 ∣ak∣∥P ∥infin It suffices to have the identity

(45) for P = ep minusN le p le N

2πip = eprimep(0) =2N

sumk=1

akep(2k minus 1

4N) =

2N

sumk=1

akep(k

2N)ep(minus

1

4N)

which we further rewrite as

2N

sumk=1

akep(k

2N) = 2πip ep(

1

4N) minusN le p le N (46)

Note that the first and the last equations coincide so we have 2N equations in2N variables Inverting the discrete Fourier transform in Z(2NZ) we obtain

ak = minusπi

N

N

sump=minusN

pep(2k minus 1

4N) + π(minus1)k

where the extra term appears since we insist on summing over 2N + 1 (ratherthan 2N) values of p

Exercise 48 Show that

ak =(minus1)kminus1π

2N sin2((2kminus1)π4N )

This implies that2N

sumk=1

∣ak∣ =2N

sumk=1

(minus1)kminus1ak = 2πN

where the last equality follows from (46) with p = N

20

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Exercise 49 For any trigonometric polynomial P isin span(eminusN ⋯ eN)

int1

0∣P prime(x)∣dx le 2πN int

1

0∣P (x)∣dx int

1

0∣P prime(x)∣2dx le 4πN 2int

1

0∣P (x)∣2dx

Now we can prove Theorem 45

Proof of Theorem 45 Let P2k be a trigonometric polynomial of degree le 2k suchthat ∥f minus P2k∥infin le A2minusak Then

∣f(x)minus f(y)∣ le 2A

2ak+ ∣P2k(x)minusP2k(y)∣ le

2A

2ak+

k

sumj=1

∣(P2j minusP2jminus1)(x)minus (P2j minusP2jminus1)(y)∣

(Please convince yourself that the dyadic decomposition is needed ie if we di-rectly apply the Bernstein inequality to estimate ∣P2k(x)minusP2k(y)∣ the conclusionwe get is much weaker than the claimed (44))

The j-th term in the sum is bounded using Bernsteinrsquos inequality

∣(P2j minus P2jminus1)(x) minus (P2j minus P2jminus1)(y)∣ le ∣x minus y∣∥(P2j minus P2jminus1)prime∥infinle ∣x minus y∣ times 2π2j times 2A2minus(jminus1)a

whence

∣f(x) minus f(y)∣ le CA [2minuska + ∣x minus y∣2(1minusa)k] C = 2π

21minusa minus 1

Choosing k so that 2minusk le ∣x minus y∣ le 2minusk+1 we obtain

∣f(x) minus f(y)∣ le C prime∣x minus y∣a C prime = C(21minusa + 1)

Remark 410 As we see the relation between the continuity properties of fand the decay of EN is rather tight This can be compared with the muchlooser connection between the continuity properties of f and the decay of itsFourier coefficients If f is continuous the Fourier coefficients tend to zero bythe RiemannndashLebesgue lemma (Lemma 311) no quantitative estimate can bemade in general (Exercise 316) On the other hand the Fourier coefficients ofa discontinuous function can decay as fast as 1p (see (33)) though not muchfaster

Exercise 411 Let (φp)pisinZ be a sequence of complex numbers

1 If sump ∣φp∣ ltinfin then

f(x) =sumpisinZφpep(x)

is a continuous function

2 If supp ∣φp∣∣p∣1+a ltinfin for some a isin (01] then f(x) is a-Holder ie ωf(δ) leCδa

21

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Fourier series of analytic functions Here is another constructive character-isation In this case there is no major difference between the rates of decay ofEN and of the Fourier coefficients

Proposition 412 Let f ∶ T rarr C and let ε gt 0 Then the following are equiva-lent

(1) f admits an analytic extension to ∣Iz∣ lt ε

(2) lim supprarrplusmninfin

1∣p∣ log ∣f(p)∣ le minus2πε

(3) lim supNrarrinfin

1N logEN(f) le minus2πε

Proof (1) OcircrArr (2) Choose εprime isin (0 ε) Then

f(p) = int1

0f(x minus iεprime)ep(minus(x minus iεprime))dx

(why) and this expression is bounded in absolute value by C(εprime) exp(minus2πpεprime)Similarly ∣f(p)∣ le C(εprime) exp(+2πpεprime)

(2) OcircrArr (1) Let εprime isin (0 ε) Choose εprimeprime isin (εprime ε) then the Fourier coefficients off admit the bound

∣f(p)∣ le C(εprimeprime) exp(minus2π∣p∣εprimeprime) whence the series

infin

sump=minusinfin

f(p)ep(z)

of analytic functions converges uniformly in the strip ∣Iz∣ lt εprime and hence definesan analytic function there This is true for any εprime isin (0 ε) whence the claim

(2) OcircrArr (3) For εprime isin (0 ε)

∣f(p)∣ le C(εprime) exp(minus2π∣p∣εprime)

hence

∥f minusN

sump=minusN

f(p)ep∥infin le sum∣p∣geN

C(εprime) exp(minus2π∣p∣εprime) le C prime(εprime) exp(minus2πNεprime)

(3) OcircrArr (2) Let εprime isin (0 ε) If for any N one can find PN for which

∥f minus PN∥infin le C(εprime) exp(minus2πNεprime)

then definitely

∣f(p)∣ = ∣f(p) minus P∣p∣minus1(p)∣ le ∥f minus P∣p∣minus1∥infin le 2πC(εprime) exp(minus2πNεprime)

22

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Exercise 413 Let f ∶ T rarr C be a function satisfying the equivalent conditionsof Proposition 412 with some ε gt 0 Then for any εprime isin (0 ε)

∣ 1

N

Nminus1

sumk=0

f(kN) minus int1

0f(x)dx∣ le C(εprime) exp(minus2πNεprime)

ie the Riemann sums converge exponentially fast

42 The heat equation

This was historically the first application of Fourier series (the name of the trea-tise of Fourier is ldquoTheorie analytique de la chaleurrdquo) The partial differentialequation

partu(t x)partt

= κ2

part2u(t x)partx2

(47)

where κ gt 0 is a parameter describes the propagation of heat in a one-dimensionalmedium If u(t x) is the temperature at time t at position x of a long rod theequation says that the rate of increase of the temperature is proportional to thedifference between the average temperature in an infinitesimal neighbourhoodof x and the temperature at x (in this case κ is called the thermal diffusivity)More formally (47) is a limit of the difference equation

u(t +∆t x) minus u(t x) = κ [u(t x +∆x) + u(t x minus∆x)2

minus u(t x)] (48)

as ∆t∆x rarr +0 ∆t = ∆x2 The left-hand side of (48) is the increment inthe temperature at x while the right-hand side is proportional to the differencebetween the average temperature in the neighbourhood of x and the temperatureat x That is the temperature at x evolves in the direction of the local average

For (47) as well as for (48) one needs to impose an initial condition

u(0 x) = u0(x) (49)

We shall assume that the rod is circular ie we impose periodic boundaryconditions u(t x + 1) = u(t x) and we set κ = 1 (since the case of general κ canbe recovered by scaling) Expand u(t x) in a Fourier series

u(t x) siminfin

sump=minusinfin

u(t p)ep(x)

The sign ldquosimrdquo is there to remind us that we do not discuss convergence yet

Exercise 414 If f ∶ T rarr C is differentiable and f prime is piecewise continuous thenf prime(p) = 2πipf(p)

23

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Using this formula we rewrite (47) as an equation on u

partu(t p)partt

= 1

2(minus4π2p2)u(t p) (410)

Note that the p-th equation of (410) only involves u(sdot p) ie we have traded apartial differential equation for an infinite list of (uncoupled) ordinary differentialequations These are solved exactly

u(t p) = u(0 p) exp(minus2π2p2t)

and we finally have

u(t x) siminfin

sump=minusinfin

u(0 p) exp(minus2π2p2t + 2πipx)

sim int1

0u(0 y)

infin

sump=minusinfin

exp(minus2π2p2t + 2πip(x minus y))dy

= [Pt lowast u(0 sdot)] (x)

(411)

where

Pt(x) =infin

sump=minusinfin

exp(minus2π2p2t + 2πipx) (412)

is called the heat kernelThe discussion so far has been purely formal However having the answer

(411) at hand we can justify it

Proposition 415 Let u0 isin C(T) Then

u(t x) =⎧⎪⎪⎨⎪⎪⎩

u0(x) t = 0

(Pt lowast u0)(x) t gt 0

defines the unique function in C([0infin)timesT)capCinfin((0infin)timesT) which satisfies theheat equation (47) for t gt 0 and x isin T with the initial condition (49)

Exercise 416 Prove the uniqueness part of Proposition 415

It is also clear that u(t sdot) = Pt lowast u0 satisfies the heat equation for t gt 0 Whatremains is to prove that u(t sdot) u0 as t rarr +0 This is straightforward for sayCinfin initial conditions u0 (why) To treat general continuous initial conditionswe need to show that Pt is a non-negative summability kernel as in the proof ofTheorem 320 (but with the t rarr +0 limit in place of N rarr infin) This is easy tocheck using the following dual representation

Lemma 417 (Jacobi identity) For any t gt 0 x isin T

Pt(x) =1radic2πt

infin

sumn=minusinfin

exp(minus(x minus n)2(2t)) (413)

24

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

minus04 minus02 0 02 04

minus6

minus4

minus2

0

x

logPt(x)

Figure 41 The logarithm of the heat kernel for t = 115 (red) the zeroth term

of (413) (blue) sum3minus3 of (412) (green)

Note that for small t the series (413) converges much faster than (412)

Proof Introduce the Cinfin periodic function

f(x) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t))

its Fourier coefficients are given by

f(p) =infin

sumn=minusinfin

exp(minus(x minus n)2(2t) minus 2πipx)dx

= intinfin

minusinfinexp(minusx2(2t) minus 2πipx)dx =

radic2πt exp(minus2π2p2t)

where the last step follows from the Gaussian integral

intinfin

minusinfinexp(minusx2(2t) + ax)dx =

radic2πt exp(a2t2) a isin C t gt 0 (414)

Exercise 418 Prove (414) It may be convenient to start with a = 0

Thus

f(x) =radic

2πtinfin

sump=minusinfin

exp(minus2π2p2t)ep(x)

as claimed

Exercise 419 Verify that Pt is a non-negative summability kernel as in the proofof Theorem 320 and compelte the proof of Proposition 415

25

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

It is instructive to inspect the solutiuon

u(t x) =infin

sump=minusinfin

u(0 p) exp(minus2π2p2t)ep(x)

The high Fourier coefficients are damped by the super-exponentially decayingterm exp(minus2π2p2t) ie by time t all the oscillations at frequencies ∣p∣ ≫ tminus12 aresmoothened out In particular u(t ) is analytic for any t gt 0 (why) In the

large t limit ut converges to its average u = int1

0 u0(x)dx If u is smooth this canbe quantified as follows

Exercise 420 Let u0 isin C(T) have a piecewise continuous derivative Then

∥ut minus u∥2 le exp(minus2π2t)∥u0 minus u∥2

(where u is understood as the constant function)

43 Toeplitz matrices and the Szego theorem

Convolution operators We start with a subject close to the initial point ofthese lecture notes (Example 11 and Exercise 23) If r = (rp)pisinZ is a two-sidedsequence of complex numbers we construct the infinite matrix

A[r] = (rpminusq)infinpq=minusinfin =

⎛⎜⎜⎜⎜⎜⎝

⋱ ⋱ ⋱⋱ r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 rminus2

r1 r0 rminus1 ⋱

⎞⎟⎟⎟⎟⎟⎠

representing convolution with the sequence (rp)pisinZ This matrix commutes withshifts therefore Fourier transform should help us diagonalise it And indeed ifwe apply it to the vector e(x) = (ep(x))pisinZ we get

(A[r]e(x))p =sumqisinZrpminusqeq(x) =sum

qisinZrpminusqeqminusp(x)ep(x) =

⎡⎢⎢⎢⎢⎣sumqisinZrqeq(minusx)

⎤⎥⎥⎥⎥⎦e(x)p

That is for each x isin T e(x) is a kind of eigenvector and the expression r(minusx)where

r(x) =sumqisinZrqeq(x)

is the corresponding eigenvalueNaturally this should not be taken too literally To speak of eigenvectors and

eigenvalues one first needs to construct an operator acting on some space Wefocus on the Hilbert space `2(Z) of two-sided sequences φ = (φp)pisinZ with

∥φ∥22 =sum

p

∣φp∣2 ltinfin

26

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Then A[r] acts on this space in a natural way

(A[r]φ)p =infin

sumq=infin

rpminusqφq =infin

sumq=minusinfin

rqφpminusq = (r lowast φ)p

To ensure that A[r] maps `2(Z) to itself we assume that r is summable

∥r∥1 =infin

sump=minusinfin

∣rp∣ ltinfin (415)

so that by CauchyndashSchwarz

∣⟨r lowast φψ⟩∣ = ∣sumpsumq

rqφpminusqψp∣ lesumq

∣rq∣sump

∣φpminusqψp∣ le ∥r∥1∥φ∥2∥ψ∥2 (416)

whence ∥A[r]∥ = ∥A[r]∥2rarr2 le ∥r∥1 This is great but unfortunately our wannabe-eigenvectors e(x) do not lie in the space

Still we would like to be able to compute the matrix elements of functionsof A[r] eg (A[r]5)00 We now try to make rigorous the idea that A[r] isdiagonalised by the Fourier transform

First observe that A[r]00 = r0 = int r(x)dx Next A[r]2 = A[r lowast r] therefore(A[r]2)00 = int r(x)2dx Similarly (A[r]2)pq = int r(x)2eqminusp(x)dx Finally we ob-tain

Proposition 421 Let r be a summable sequence of complex numbers Thenfor any polynomial u(λ) and any p q isin Z

u(A[r])pq = int1

0u(r(x))eqminusp(x)dx = ((u r)and)pminusq (417)

This fact can be concisely written as

(Flowastu(A[r])Ff)(x) = u(r(minusx))f(x) (418)

Observe that the right-hand side of (417) (or of (418)) makes sense for anyu isin C(R) Therefore we can view (417) as a definition of the operator u(A[r])Exercise 422 The map u ↦ u(A[r]) = A[(u r)and] from C(R) to the space ofbounded operators on `2(Z) equipped with norm topology is continuous

Now we discuss an application for which polynomials suffice

Exercise 423 Let T be the transition matrix of the random walk on Z ieTij = 1

21∣iminusj∣=1 Prove that

1 suminfink=0[T k]00 = suminfin

k=0 int12

minus 12

cosk(2πx)dx = infin ie the number of returns of the

random walk to its starting point has infinite expectation

27

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

2 with probability one a random walk on Z returns to its starting pointinfinitely many times (The same is true on Z2 but not on Z3 ndash why)

Finally we comment on the condition ∥r∥1 ltinfin which is a special case of

Exercise 424 (Schurrsquos test) A matrix A = (apq)infinpq=minusinfin such that

α = suppsumq

∣apq∣ ltinfin β = supqsump

∣apq∣ ltinfin

defines a bounded operator on `2(Z) and moreover ∥A∥ leradicαβ

Exercise 425 Does there exist r = (rp)pisinZ such that ∥r∥1 =infin and yet ∥A[r]∥ ltinfin

Toeplitz matrices Our next goal is to study the finite restrictions of A[r]namely the N timesN matrices

(AN[r])pq = rpminusq 1 le p q le N (419)

A matrix of this form is called a Toeplitz matrix Our goal is to find the asymp-totic behaviour of detAN[r] The answer turns out to be

limNrarrinfin

[detAN[r]]1N = exp [int

1

0log r(x)dx] (420)

where the right-hand side is the geometric mean of r (at least if r ge 0) Thisanswer is natural for the following reason

detAN[r] = exp tr logAN[r] = expN

sump=1

(logAN[r])pp (421)

if we would be able to replace AN[r] with A[r] in the last equality we would getby (417)

(421) asymp expN

sump=1

(logA[r])pp = exp [N int1

0log r(x)dx] (422)

as claimed However the justification of the approximation (logAN[r])pp asymp(logA[r])pp is a delicate matter having to do with the influence of the boundaryand we shall proceed along different lines We start with a digression

Szego theorem For an integrable w ∶ Trarr R+ denote by

G(w) = exp [int1

0logw(x)dx]

the geometric mean of w understood to be zero if the integral diverges to minusinfin

28

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Theorem 426 (Szego) If w ∶ Trarr R+ is piecewise continuous then

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣eminus1(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = G(w)

Example 427 If w equiv 1 the left-hand side is obviously equal to one (the infimumis attained when all the coefficients are zero) and so is the right-hand side Onthe other hand if w vanishes on an open interval the right-hand side equalszero (in fact in this case the convergence to zero is exponentially fast) You arewelcome to prove the following for any open interval I sub T any f isin C(T) canbe approximated by linear combinations of (ep)pge0 uniformly on the complementof I

The conclusion of the Szego theorem can be amplified as follows

Exercise 428 If w ∶ T rarr R+ is piecewise continuous then G(w) = 0 if and onlyif for any piecewise continuous φ ∶ Trarr C

limNrarrinfin

infc0⋯cNminus1isinC

int1

0∣φ(x) minus

Nminus1

sump=0

cpep(x)∣2w(x)dx = 0

(You are welcome to use the theorem if needed)

We also note the following corollaryrestatement (cf (420))

Exercise 429 In the setting of the Szego theorem we have for AN[w] from(419)

limNrarrinfin

[detAN[w]]1N = G(w)

Now we prove the theorem We need a few basic properties of subharmonicfunctions (see any textbook in complex analysis eg Ahlfors [1978])

Definition 430 Let D sub R2 be a domain An upper semicontinuous functionu ∶ D rarr R cup minusinfin is called subharmonic if for any (x0 y0) isin D and any 0 lt r ltdist((x0 y0) partD)

u(x0 y0) le int1

0u(x0 + r cos(2πx) y0 + r sin(2πx))dx (423)

Remark 431

1 It is sufficient to impose the condition (423) for 0 lt r lt r0(x0 y0) Anotherequivalent property is that ∆u ge 0 in distribution sense ie

intDu(x y)∆φ(x y)dxdy ge 0 (424)

for any smooth non-negative test function φ which is compactly supportedin D

29

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

2 If F (z) is analytic in D then u(x y) = log ∣F (x + iy)∣ is subharmonicFurther if F does not have zeros in D then u is harmonic meaning that(423) (or (424)) is always an equality Both of these facts follow from theidentity

intD

log ∣F (x + iy)∣ (∆φ)(x y)dxdy = 2π sumF (x+iy)=0

φ(x y)

valid for any φ which is smooth and compactly supported in D (with mul-tiplicity taken into account on the right-hand side)

Proof of Theorem 426 We shall prove the following identity (equivalent to theclaimed one)

limNrarrinfin

infc1⋯cNisinC

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx = G(w)

We start with the inequality ge By Jensenrsquos inequality

logint1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge int1

0log [∣1 minus

N

sump=1

cpep(x)∣2w(x)]dx

= int1

0log ∣1 minus

N

sump=1

cpep(x)∣2dx + int1

0logw(x)dx

(425)

From (423) applied to u(x y) = log ∣1 minussumNp=1 cp(x + iy)p∣ the first integral is

= 2int1

0u(cos(2πx) sin(2πx))dx ge u(00) = 0 (426)

whence

int1

0∣1 minus

N

sump=1

cpep(x)∣2w(x)dx ge G(w)

Now we prove le To saturate both inequalities (425) and (426) we needto find P (z) = 1 minussumNp=1 cpz

p such that ∣P (e2πix)∣2w(x) is approximately constantand also P has no zeros in the interior of the unit disc We can assume withoutloss of generality that w is bounded away from zero Indeed let wδ = max(w δ)then

limδrarr+0int logwδ(x)dxrarr int logw(x)dx

(why) whereas

int ∣P (x)∣2wδ(x)dx ge int ∣P (x)∣2w(x)dx

30

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Therefore we assume that there exists M ge δ gt 0 such that 0 lt δ le w le M LetQη be a trigonometric polynomial such that ∣Qη minus 1w∣ le η except for a union ofintervals of total length le η and such that Mminus1 minus η le Qη le δminus1 + η Then

∣int Qη(x)w(x)dx minus 1∣ leM(1 + δminus1 + η)η rarr 0 as η rarr +0

Thus we have as η rarr +0

logint Qη(x)w(x)dx le int Qη(x)w(x)dx minus 1rarr 0

int log(Qη(x)w(x))dx ge log(1 minusMη) minus η log(M(δminus1 + η))rarr 0

hence Qηw almost saturates Jensenrsquos inequality

logint Qη(x)w(x)dx minus int log(Qη(x)w(x))dxrarr 0 as η rarr +0 (427)

Now we need

Lemma 432 (RieszndashFejer) Let Q(x) be a trigonometric polynomial which isnon-negative on T Then there exists a polynomial P (z) with no zeros in D suchthat Q(x) = ∣P (e2πix)∣2 on T

Applying the lemma to Qη we get a polynomial Pη all the zeros of which lie

outside D Set Pη(z) = Pη(z)Pη(0) Then (427) implies that

logint ∣Pη(e2πix)∣2w(x)dx minusG(w)

= logint ∣Pη(e2πix)∣2w(x)dx minus int log(∣Pη(e2πix)∣2w(x))dxrarr 0 as η rarr +0

as claimed Now we proceed to

Proof of Lemma 432

Let Q(x) = sumNp=minusN cpep(x) so that cminusN ne 0 Then R(z) = sumNp=minusN cpzN+p isan algebraic polynomial that satisfies R(0) ne 0 and Q(x) = ∣R(e2πix)∣ Due to

the relation R(1z) = zminus2NR(z) the zeros of R on the unit circle are of evenmultiplicity whereas the roots not on the unit circle come in pairs (z1z)Therefore

R(z) = cprodj

(z minus zj)(z minus 1zj) where c gt 0 ∣zj ∣ ge 1

and we can setP (z) =

radiccprod

j

(z minus zj)

31

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Predictability of Gaussian processes Here is an application Let (Xj)jisinZ be astationary Gaussian process with EXj = 0 It is called predictable if

limNrarrinfin

infc0⋯cNminus1

∣X1 minusNminus1

sump=0

cpXminusp∣2 = 0

The Szego theorem allows to derive a necessary and sufficient condition for pre-dictability we shall discuss a special case

The covariancesrpminusq = EXpXq

always form a positive semidefinite matrix (rpminusq)pqisinZ ⪰ 0 ie

sumpq

cprpminusqcq = E∣sum cpXp∣2 ge 0 (428)

for any complex coefficients cp Vice versa any sequence (rp)pisinZ for which(rpminusq)pq ⪰ 0 corresponds to a stationary Gaussian process Here is a way toconstruct such sequences

rp = w(p) = int1

0eminus2πixw(x)dx w ge 0 (429)

Exercise 433 Any sequence rp of the form (429) satisfies (rpminusq)pq ⪰ 0

The function w(x) is called the spectral function of the process (Xp)pisinZ

Exercise 434 Let w ∶ Trarr R+ be piecewise continuous The stationary Gaussianprocess (Xj) defined by

EXpXq = w(p minus q)is predictable if and only if G(w) = 0

Remark 435 The construction of stationary Gaussian processes described aboveis almost the most general one if we replace the piecewise continuous functionw(x) with a non-negative measure on the circle we get the general form ofa sequence (rp) satisfying (rpminusq) ⪰ 0 (see Exercise 457 below) leading to thegeneral form of a stationary Gaussian process

See Grenander and Szego [1984] or Dym and McKean [1972] for the generalversion of the Szego theorem and the general version of Exercise 434

44 Central limit theorem

Let SN = S1 + ⋯ + SN be independent identically distributed random variableswith EX1 = a and E(X1 minus a)2 = σ2 The LevyndashLindeberg Central Limit Theoremasserts that

SN minus aNσradicN

rarr N(01)

32

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

in distribution ie

forallα le β limNrarrinfin

PSN minus aNσradicN

isin [αβ] = intβ

α

1radic2πeminuss

22ds (430)

The simplest case (studied by de Moivre and Laplace) is whenXj are Bernoulli(12)distributed

Xj =⎧⎪⎪⎨⎪⎪⎩

minus1 with probability 12+1 with probability 12

In this case

P(SN = k) =⎧⎪⎪⎨⎪⎪⎩

2minusN( N(Nminusk)2) minusN le k le N N minus k isin 2Z

0

and Stirlingrsquos approximation n = (1 + o(1))radic

2πe(ne)n implies that for anyδ lt 14 one has

P(SN = k) = (1 + o(1)) 2radic2πN

eminusk2(2N) ∣k∣ le N 12+δ N minus k isin 2Z (431)

with a uniform o(1) term

Exercise 436 Check that (431) implies (430)

The combinatorial approach does not easily extend to other random variablesHowever the following result is quite general

Theorem 437 (Gnedenko) Let Xj be independent copies of an integer-valuedrandom variable X such that EX = a and E(X minus a)2 = σ2 If suppX does not liein any sublattice of Z in other words if the differences k minus kprime ∣ k kprime isin suppXgenerate Z as a group then

limNrarrinfin

supkisinZ

radicN ∣P(SN = k) minus 1radic

2πσ2Nexp [minus(k minus a)2

2σ2N]∣ = 0 (432)

Exercise 438 Explain the inconsistency between (431) and (432)

Exercise 439 Check that (432) implies (430)

The proof uses characteristic functions Let π(k) = P(X = k) and πN(k) =P(SN = k) Then

πN(k) = sumk1+⋯+kN=k

π(k1)⋯π(kN) = (π lowast⋯ lowast π)acutesup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1cedilsup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1sup1para

N times

(k)

ie πN is the N -fold convolution power of π Set

π(x) =sumkisinZ

π(k)ek(x) πN(x) =sumkisinZ

πN(k)ek(x)

so that (π)and = π and (πN)and = πN Then πN = (π)N

33

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Exercise 440 Let π be a probability distribution on the integers

1 π isin C(T)

2 if for some m isin N the m-th absolute moment sumk π(k)∣k∣m lt infin thenπ isin Cm(T)

In our case π isin C2(T) and

π(0) = 1 πprime(0) = 2πia πprimeprime(0) = minus4π2(a2 + σ2) (433)

ieπ(x) asymp 1 + 2πiax minus 2π2(a2 + σ2)x2 asymp exp(2πiax minus 2π2σ2x2) (434)

and one can hope that

πN(k) asymp int12

minus12exp(2πi(aN minus k)x minus 2π2Nσ2x2)dx (435)

For large N the integrand is very small outside (minus1212) hence we can furtherhope that

(435) asymp intinfin

minusinfinexp(2πi(aN minusk)xminus2π2Nσ2x2)dx = 1radic

2πexp(minus(kminusaN)2(2Nσ2))

Here the last equality follows from (414) To justify this argument we first needto handle x which are not very close to 0 For such x (434) and (435) are notvalid however we can hope that πN is small

Example 441 For Bernoulli(12) random variables π(x) = cos(2πx) hence∣πN(12)∣ = 1 whereas the right-hand side of (435) is exponentially small

Luckily Bernoulli(12) random variables do not satisfy the assumptions of thetheorem (see Exercise 438 above) More generally we have

Lemma 442 Let π be a probability distribution on the integers such that thedifferences

k minus kprime ∣k kprime isin suppπgenerate Z (as a group) Then maxxisinT ∣π(x)∣ is uniquely attained at x = 0

Proof Suppose ∣π(x)∣ = 1 for some x ne 0 Then

∣sumkisinZ

π(k)ek(x)∣ =sumkisinZ

π(k)∣ek(x)∣

ie ek(x) is constant on suppπ whence ekminuskprime(x) = 1 for any k kprime isin suppπ Thisimplies that x is rational x = ab and all the differences k minus kprime lie in bZ

34

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Proof of Theorem 437 According to (433) and the Taylor expansion with Peanoremainder

∣π(x) minus exp(2πiax minus 2π2σ2x2)∣ le x2ω(x) ∣x∣ le 1

2

where ω ∶ [012] rarr R+ is non-decreasing and tends to 0 as the argument ap-proaches zero This implies

∣ log π(x) minus (2πiax minus 2π2σ2x2)∣ le 2x2ω(x) ∣x∣ le c1

where c1 is small enough to ensure that Rπ gt 0 in this region whence

∣ log πN(x) minusN(2πiax minus 2π2σ2x2)∣ le 2Nx2ω(x) ∣x∣ le c1

Now we write

∣πN(k) minus 1radic2πσ2N

eminus(kminusa)2(2σ2N)∣

= ∣int12

minus12πN(x)ek(x)dx minus int

infin

minusinfineminusN(2πiaxminus2π2σ2x2)ek(x)dx∣ le Aprime +Aprimeprime +Aprimeprimeprime

where

Aprime = intR

radicN

minusRradicN

∣πN(x) minus eminusN(2πiaxminus2π2σ2x2)∣dx

Aprimeprime = int RradicNle∣x∣le 1

2

∣πN(x)∣dx

Aprimeprimeprime = int RradicNle∣x∣

∣eminusN(2πiaxminus2π2σ2x2)∣dx = int RradicNle∣x∣eminus2Nπ2σ2x2dx

Exercise 443 Show thatAprime le CR2ω(RradicN)

radicN Aprimeprime+Aprimeprimeprime le C

RradicN

exp(minus2π2σ2R2)and complete the proof of the theorem

45 Equidistribution modulo one

The following is based on the work of Weyl We mostly follow Montgomery[2014] A sequence (an)nge1 of real numbers is said to be equidistributed moduloone if for any arc I sub T one has

limNrarrinfin

1

N1 le n le N ∶ an +Z isin I = ∣I ∣ (436)

ie the number of points in each arc is asymptotically proportional to the lengthof the arc

35

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Exercise 444 Let X1X2⋯ be a sequence of independent random variables uni-formly distributed in [01] Then almost surely (Xn) is equidistributed moduloone

Exercise 445 A sequence (an) is equidistributed modulo one if and only if forany f isin C(T)

limNrarrinfin

1

N

N

sumn=1

f(an) = int1

0f(x)dx

Corollary 446 (Weylrsquos criterion) A sequence (an) is equidistributed moduloone if and only if for any p ge 1

limNrarrinfin

1

N

N

sumn=1

ep(an) = 0 (437)

The convenient feature of Weylrsquos criterion (437) is that it is a countable listof conditions which are often not hard to check Informally the p-th conditionis responsible for equidistribution on scale asymp 1pExercise 447 Solve Exercise 444 using Weylrsquos criterion

Here is a more interesting example

Claim 448 If α isin R ∖Q the sequence (an = αn)nge1 is equidistributed moduloone

The conclusion clearly fails if α is rational (why)

Proof For each p ne 0

N

sumn=1

ep(an) =ep(α)(1 minus ep(Nα))

1 minus ep(α)(438)

is bounded (as N varies) since the absolute value of the numerator is at mosttwo and the denominator does not vanish Therefore the left-hand side of (437)is O(1N)

Remark 449 The better α is approximated by rationals the slower is the con-vergence in (436) A quantitative version of Weylrsquos criterion was found by Erdosand Turan see Montgomery [2014]

Exercise 450 Let f isin C(T) Then for any α isin R ∖Q

1

N

N

sumn=1

f(x + nα) int1

0f(x)dx N rarrinfin

Theorem 451 (Weyl) Let P isin R[x] be a polynomial with at least one irrationalcoefficient (next to a power ge 1) Then (P (n))nge1 is equidistributed modulo one

36

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Exercise 452 It is sufficient to prove the theorem for polynomials with irrationalleading coefficient

Proof We argue by induction on degP Claim 448 takes care of the inductionbase The induction step follows from the next theorem

Theorem 453 (van der Corput) Let (an)nge1 be a sequence of real numbersIf for any h ge 1 (an+h minus an)nge1 is equidistributed modulo one then (an)nge1 isequidistributed modulo one

Exercise 454 Is the converse true

The claim is plausible since

∣ 1

N

N

sumn=1

ep(an)∣2

= 1

N

N

sumh=1

( 1

N

N

sumn=1

ep(a(N)n minus a(N)

n+h)) (439)

where aNn = an for 1 le n le N and then continues periodically Now

∣N

sumn=1

ep(a(N)n minus a(N)

n+h) minusN

sumn=1

ep(an minus an+h)∣ le 2h

therefore for each fixed h the expression in the large parentheses of (439) tendsto zero The problem is that we need to sum over all 1 le h le N whereas theassumptions of the theorem imply no uniformity in h The argument of van derCorput presented below allows to restrict the summation to a finite range of h

Lemma 455 (van der Corput) For any z1⋯ zN isin C and any H ge 1

H2 ∣N

sumn=1

zn∣2

leH(N +H minus 1)N

sumn=1

∣zn∣2 + 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

Proof Set zn = 0 for n ge N + 1 and for n le 0 Then

HN

sumn=1

zn =Hminus1

sumr=0

N+Hminus1

sumn=1

znminusr =N+Hminus1

sumn=1

Hminus1

sumr=0

znminusr

whence

H2 ∣N

sumn=1

zn∣2

le (N +H minus 1)N+Hminus1

sumn=1

∣Hminus1

sumr=0

znminusr∣2

(Jensen) (440)

= (N +H minus 1)N+Hminus1

sumn=1

Hminus1

sumrs=0

znminusrznminuss (441)

le (N +H minus 1)HN

sumn=1

∣zn∣2 (442)

+ 2(N +H minus 1)Hminus1

sumh=1

(H minus h) ∣Nminush

sumn=1

zn+hzn∣

37

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Proof of Theorem 453 Let

ANp =1

N

N

sumn=1

ep(an) ANph =1

N

N

sumn=1

ep(an+h minus an)

Fix p ge 1 we know that ANph rarr 0 for any h ge 1 and we need to show thatANp rarr 0 Applying van der Corputrsquos lemma to zn = ep(an) we obtain

H2N 2 ∣ANp∣2 leH(N +H minus 1)N + 2(N +H minus 1)Hminus1

sumh=1

(H minus h)(N minus h)∣ANminushph∣

whence for large N

∣ANp∣2 le2

H+ 3

H

Hminus1

sumh=1

∣ANminushph∣

It remains to let N rarrinfin and then H rarrinfin

Exercise 456 The sequence (radicn)nge1 is equidistributed modulo 1

Fourier series of functionals Weylrsquos criterion is not a statement about Fourierseries in the generality of our discussion so far One way to embed it into theharmonic analysis framework is as follows Let micro ∶ C(T) rarr C be a continuousfunctional One may think eg of the functionals

νg ∶ f ↦ int1

0fgdx δx0 ∶ f ↦ f(x0) ⋯ (443)

Define the Fourier coefficients micro(p) = micro(eminusp) Then

microN = 1

N

N

sumn=1

δan rarr ν1 (444)

if and only if microN(p) rarr ν1(p) = 0 for any p isin N (what about p = 0 and negativep) The convergence (443) of functionals on C(T) is equivalent (in this case) tothe equidistribution of (an) cf Exercise 445

The functionals which we considered in this section had a special positivityproperty which we highlight below

Exercise 457 Let micro ∶ C(T) rarr C be a continuous functional The following areequivalent

1 micro sends (pointwise) non-negative functions to non-negative numbers

2 micro sends (pointwise) non-negative trigonometric polynomials to non-negativenumbers

38

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

3 the matrix (micropminusq)pqisinZ is non-negative semi-definite (ie any principal sub-matrix of finite size is non-negative semi-definite)

Exercise 458 Let (rp)pisinZ be a sequence such that (rpminusq)pqisinZ is non-negative semi-definite Then there exists micro satisfying the equivalent conditions of Exercise 457such that r = micro (cf Remark 435)

One can go further and consider functionals on smaller functional spaces con-taining ep eg the space Cinfin(T)Exercise 459 Compute the Fourier coefficients of the functional f ↦ f primeprime(x0)

5 Fourier transform

51 Introduction

Now we consider the group R Consider the characters

ep(x) = exp(2πipx) p isin R

Note that we insist that p isin R (these are singled out for example by the require-ment that a character be a closed map or by insisting that the image lie in theunit circle) For a nice (piecewise continuous and absolutely integrable) functionf(x) set

(Ff)(p) = f(p) = intinfin

minusinfinf(x)ep(x)dx

Similarly for a nice φ(p) set

(Flowastφ)(x) = φ(x) = intinfin

minusinfinφ(p)ep(x)dp

Similarly to Section 32

intR

minusRf(p)ep(x)dp = int

infin

minusinfinDR(x minus y)f(y)dy = (DR lowast f)(x)

where the Dirichlet kernel is now given by

DR(y) = intR

minusRep(y)dp =

sin(2πRy)πy

(51)

Exercise 51 Prove the RiemannndashLebesgue lemma if f is piecewise continuousand absolutely integrable then f(p)rarr 0 as prarr plusmninfin

39

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Exercise 52 Let a isin (01] If f is piecewise continuous and absolutely integrableand if

∣f(x) minus f(x0)∣ le C ∣x minus x0∣a

for all x in a neighbourhood of a certain x0 isin R then (DR lowastf)(x0)rarr f(x0) andmoreover

limR+Rminusrarr+infin

intR+

minusRminusf(p)ep(x0)dp = f(x0)

Proposition 53 Let f ∶ Rrarr C be bounded absolutely integrable and uniformlycontinuous Then

1

R intR

0drint

r

minusrf(p)ep(x)dp f(x)

Proof The proof follows the lines of Section 33 The starting point is the identity

1

R intR

0drint

r

minusrf(p)ep(x)dp = (SR lowast f)(x) SR(x) =

1

R

sin2(πRx)(πx)2

52 Fourier transform in Schwartz space

The Schwartz space S(R) is defined as

S(R) = f isin Cinfin(R) ∣forallk ` ge 0 ∣∣∣f ∣∣∣kl = supxisinR

∣x∣k∣f (`)(x)∣ ltinfin

Observe that every f isin S(R) is absolutely integrable hence f is well-definedWe also note that S(R) is closed under differentiation and multiplication by x(why)

Proposition 54 The map F ∶ f ↦ f is a bijection S(R)rarr S(R) and Flowast is itsinverse

Proof In view of Exercise 52 it suffices to check that F(S(R)) sub S(R) To thisend we compute

∣∣∣f ∣∣∣k` = suppisinR

∣p∣k∣f (`)(p)∣

= suppisinR

∣p∣k∣(minus2πip)`f)and(p)∣

= (2π)`minusk suppisinR

∣( dk

dxkx`f)and(p)∣

le (2π)`minusk∣∣∣ dk

dxk(x`f)∣∣∣00

40

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Exercise 55 (cf Exercise 418)

f(x) = exp(minusx2

2t+ ax)OcircrArr f(p) =

radic2πt exp((2πip minus a)2t2)

Plancherel indentity

Proposition 56 (Plancherel) Let f isin S(R) Then

int ∣f(p)∣2dp = int ∣f(x)∣2dx

We shall use the following converse to Exercise 321

Exercise 57 Let u ∶ R+ rarr R+ be a non-decreasing function If

limRrarrinfin

1

R intR

0u(r)dr = L isin [0+infin]

then alsolimrrarrinfin

u(r) = L

Proof of Proposition 56 The integral intrminusr ∣f(p)∣2dp is non-decreasing in r there-

fore it suffices to show that

limRrarrinfin

1

R intR

0int

r

minusr∣f(p)∣2dp = int ∣f(x)∣2dx (52)

Interchanging the limits (by what rights) in the identity

intr

minusr∣f(p)∣2dp = int

r

minusr[int f(x)ep(x)dxint f(y)ep(y)dy]dr

we obtain

intr

minusr∣f(p)∣2dp = int dyf(y)int dxf(x)Dr(y minus x)

whenceLHS of (52) = int dyf(y)int dxf(x)Sr(y minus x)

By Proposition 53

int dxf(x)Sr(y minus x) f(y) r rarr +infin

henceLHS of (52)rarr int ∣f(x)∣2dx

Exercise 58 Show that for any f g isin S(R)

int f(p)g(p)dp = int f(x)g(x)dx

Remark 59 The Plancherel identity allows to extend the Fourier transform toan isometry of the space of (Lebesgue-) square-summable functions

41

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

53 Poisson summation

Proposition 510 Let f isin S(R) Then

infin

sump=minusinfin

f(p) =infin

sumn=minusinfin

f(n)

Proof Let g(x) = sumnisinZ f(x + n) Then g isin Cinfin(T) and

g(p) =sumnisinZint

n+1

nf(x)ep(x)dx = int

infin

minusinfinf(x)ep(x)dx = f(p) p isin Z

where the hat on the left-hand side stands for the Fourier coefficients and onthe right-hand side mdash for the Fourier transform Thus

sumnisinZ

f(n) = g(0) =sumpisinZg(p) =sum

pisinZf(p)

One application is the Jacobi identity which we have already discussed inSection 42 (using a similar argument) Here we employ the following notationthe Jacobi theta function is defined as

θ(z) =infin

sumn=minusinfin

exp(πin2z) Iz gt 0

Exercise 511 θ(minus1z) =radicminusiz θ(z) where the principal branch of the square

root is taken

Digression Riemann ζ-fiunction We now use the Jacobi identity to derivethe functional equation for the Riemann ζ-function For Rs gt 1 denote

ξ(s) = πminuss2Γ(s2)ζ(s) = πminuss2 times [intinfin

0us2eminusu

du

u] times [

infin

sumn=1

nminuss]

Theorem 512 ξ(s) admits a meromorphic continuation to C with simple polesat s = 01 and no other poles It satisfies the functional equation ξ(1 minus s) = ξ(s)

We follow Green [2016]

Proof We first observe that for s gt 1

ξ(s) = intinfin

0(θ(iu) minus 1

2)u s

2du

u (53)

Indeed

intinfin

0exp(minusπn2u)u s

2du

u= πminus s2Γ(minuss2)nminuss

42

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Now we split the integral (53) in two parts The integral from 1 to infin is alreadyan entire function therefore we keep it as it is The integral from 0 to 1 can betransformed using Exercise 511

int1

0(θ(iu) minus 1

2)u s

2du

u= 1

2 int1

0θ(iu)u s

2du

uminus 1

s

= 1

2 intinfin

1θ(minus(iu)minus1)uminus s2 du

uminus 1

s= 1

2 intinfin

1

radicuθ(iu)uminus s2 du

uminus 1

s

= 1

2 intinfin

1θ(iu)u 1minuss

2du

uminus 1

s= int

infin

1(θ(iu) minus 1

2)u 1minuss

2du

uminus 1

sminus 1

1 minus s

Thus

ξ(s) = intinfin

1(θ(iu) minus 1

2) [u s

2 + u 1minuss2 ] du

uminus 1

sminus 1

1 minus swhich enjoys the claimed properties

Now we can show that ζ has no zeros on the boundary of the critical strip

Theorem 513 The ζ-function has no zeros on the line Rs = 1

This theorem is the key ingredient in the proof of the prime number theorem

(the number of primes le x) = (1 + o(1)) x

logx xrarr +infin

See further Green [2016] or Dym and McKean [1972]

Proof Following Gorin [201112] let σ gt 1 and consider the function fσ(t) =log ∣ζ(σ + it)∣ By the Euler product formula

ζ(s) = prodprime p

(1 minus pminuss)minus1

we have

log ζ(s = σ + it) = minussump

log(1 minus pminuss) =sumpsumnge1

1

npsn=sum

psumnge1

1

npσnen log p(t)

Therefore

fσ(t) =sumpsumnge1

1

npσn[en log p(t) + eminusn log p(t)]

The key feature of this formula is that all the coefficients are positive ie

N

sumjk=1

fσ(tj minus tk)cjck ge 0

43

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

for any cj isin C tj isin R (does this remind you of (428)) Take t1 = t gt 0 t2 = 0t3 = minust and c1 = c2 = c3 = 1 We obtain

3fσ(0) + 4fσ(t) + 2fσ(2t) ge 0

As σ rarr 1 + 0 fσ(t) sim minusm(t)∣ log(σ minus 1)∣ where m(t) = +m if 1 + it is a zero ofmultiplicity m and m(t) = minusm if t is a pole of multiplicity m By Theorem 512m(0) = minus1 and m(t) ge 0 for any other t therefore m(t) le 34 whence it iszero

54 The uncertainty principle

A digression to quantum mechanics In the Schrodinger picture a state ofa system is described by a unit vector ψ isin H where H is a Hilbert space Anobservable is described by a self-adjoint operator acting on H The average ofan observable A in a state ψ is given by EψA = ⟨Aψψ⟩ The variance of A is

VarψA = ⟨A2ψψ⟩ minus ⟨Aψψ⟩2 = Eψ(A minus (EψA)1)2

One can further construct a random variable a such that for any decent functionu one has ⟨u(A)ψψ⟩ = Eu(a) The probability distribution of a is called thedistribution of A in the state ψ If ψ is an eigenvector of A with eigenvalue λthen ⟨u(A)ψψ⟩ = u(λ) hence a equiv λ If ψ is not an eigenvalue of A then thedistribution of A is non-trivial

In classical mechanics one can measure the position x and the momentum pof a (say one-dimensional) particle these are examples of classical observablesIn quantum mechanics these are reblaced by quantum observables X and P The fundamental relation is XP minusPX = ih1 where h asymp 10minus34 J sdot sec is a structuralconstant of quantum mechanics (the reduced Planck constant) It implies thatX and P can not have joint eigenvectors ie for each ψ at least one of them hasnon-zero variance

The description of quantum mechanics is unitary-invariant and does not de-pend on the choice of the underlying Hilbert space H However it is often usefulto have in mind a realisation a particularly convenient one is H = L2(R) (thespace of Lebesgue square-integrable functions) X is the operator of multiplica-tion by x and p is a multiple of the differentiation operator

(Xψ)(x) = xψ(x) (Pψ)(x) = minusihψprime(x)

Then indeed (XP minusPX)ψ = ihψ (at least this relation formally holds for polyno-mials ψ(x)) For the sake of concreteness we confine ourselves to this realisation

Exercise 514 For any f g isin S(R) (XP minus PX)f = minusihf

44

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Theorem 515 (Heisenberg uncertainty principle) For any ψ isin S(R) such that∥ψ∥ = 1

VarψX timesVarψ P ge h2

4

Proof By the commutation relation

ih = ih⟨ψψ⟩ = ⟨(XP minus PX)ψψ⟩= ⟨((X minus (EψX)1)(P minus (EψP )1) minus (P minus (EP )1)(X minus (EψX)1))ψψ⟩

Now we estimate

∣⟨(X minus (EψX)1)(P minus (EψP )1)ψψ⟩∣ = ∣⟨(P minus (EψP )1)ψ (X minus (EψX)1)ψ⟩∣le ∥(P minus (EψP )1)ψ∥∥(X minus (EψX)1)ψ∥ =

radicVarψX timesVarψ P

and we are done

Back to harmonic analysis We now return to the mathematical framework ofharmonic analysis and from now on we set h = 1

2π With this choice

FlowastPF =X

and we obtain

Corollary 516 For any f isin S(R)

infx0p0isinR

[int ∣x minus x0∣2∣f(x)∣2dx times int ∣p minus p0∣2∣f(p)∣2dp] ge∥f∥4

2

16π2 (54)

With some more work one can show that this inequality holds whenever thequantities that appear in it are finite

Fourier transform of compactly supported functions The inequality (54) is

one of many results asserting that f and f can not be simultaneously localisedHere is another one

Exercise 517 Suppose f isin S(R) is compactly supported and also f is compactlysupported Then f equiv 0

We now prove a much stronger statement

Theorem 518 (PaleyndashWiener) Let R gt 0 A function f ∶ R rarr C can beanalytically continued to an entire function satisfying the estimates

forallk ge 0 supzisinC

∣f(z)∣(1 + ∣z∣k) exp(minus2πR∣Iz∣) ltinfin (55)

if and only if f isin S(R) and supp f sub [minusRR]

45

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Proof Suppose f isin S(R) and supp f sub [minusRR] Then we can define an analyticextension

f(x + iy) = intR

minusRf(p) exp(2πip(x + iy))dp

and

∣f(x+iy)∣ le intR

minusR∣f(p)∣ exp(2π∣p∣∣y∣)dp le int

R

minusR∣f(p)∣dp exp(2πR∣y∣) = C0 exp(2πR∣y∣)

This proves the case k = 0 of (55) Applying this to the functions fk(x) = f(x)xkwe obtain

∣f(x + iy)∣∣x + iy∣k le Ck exp(2πR∣y∣) Vice versa if f satisfies (55) its restriction to R definitely lies in the Schwartzspace

f(p) = int f(x) exp(minus2πixp)dx

For p gt 0 we can deform the contour of integration to the line Iz = minusr lt 0 Weget

∣f(p)∣ le intinfin

minusinfin∣f(x minus ir)∣ exp(minus2πrp)dx le int

infin

minusinfin

C2 exp(+2πrR)1 + x2

exp(minus2πrp)dx

and for p gt R this tends to zero as r rarrinfin Similarly f(p) vanishes for p lt minusR

See eg Dym and McKean [1972] for the more delicate L2 version of this result

In fact functions with compactly supported Fourier transform share manyproperties with (trigonometric) polynomials Here is a Bernstein-type inequality

Exercise 519 If f isin S(R) and supp f sub [minusRR] then ∥f prime∥2 le 2πR∥f∥2

A similar inequality is true for the sup-norm (cf Theorem 45)

Exercise 520 Let f isin S(R) supp f sub [minusRR] Prove that

f(x) =infin

sumn=minusinfin

f(nR)sin(π(xR minus n))π(xR minus n)

Consequently if f g isin S(R) supp f g sub [minusRR] then

int f(x)g(x)dx =infin

sumn=minusinfin

f(nR)g(nR)

46

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Some notation

ep(x) exp(2πipxN) (Sections 1 2) exp(2πipx) (Sections 3ndash5)

T The circle RZf Fourier coefficient of f (Sections 1ndash4) Fourier transform of f (Section 5)

F The map f ↦ f

∥f∥ = ∥f∥2 [int ∣f(x)∣2dx]12

∥f∥infin sup ∣f ∣DN DR Dirichlet kernel (37) (51)

ωf(δ) Modulus of continuity see (42)

EN(f) the measure of approximation of f by (trigonometric) polynomials see (41)

S(R) Schwartz space

References

Lars V Ahlfors Complex analysis McGraw-Hill Book Co New York thirdedition 1978 ISBN 0-07-000657-1 An introduction to the theory of analyticfunctions of one complex variable International Series in Pure and AppliedMathematics

H Dym and H P McKean Fourier series and integrals Academic Press NewYork-London 1972 Probability and Mathematical Statistics No 14

A A Giannopoulos and V D Milman Concentration property on probabilityspaces Adv Math 156(1)77ndash106 2000 ISSN 0001-8708 doi 101006aima20001949

E A Gorin Positive-definite functions as a tool of mathematical analy-sis Fundam Prikl Mat 17(7)67ndash95 201112 ISSN 1560-5159 doi101007s10958-014-1730-5

W T Gowers Fourier analysis and Szemeredirsquos theorem In Proceedings ofthe International Congress of Mathematicians Vol I (Berlin 1998) numberExtra Vol I pages 617ndash629 1998

Ben Green Analytic number theory 2016 URL httppeoplemathsoxac

ukgreenbjpapersprimenumberspdf

Ulf Grenander and Gabor Szego Toeplitz forms and their applications ChelseaPublishing Co New York second edition 1984 ISBN 0-8284-0321-X

47

Yitzhak Katznelson An introduction to harmonic analysis Cambridge Mathe-matical Library Cambridge University Press Cambridge third edition 2004ISBN 0-521-83829-0 0-521-54359-2 doi 101017CBO9781139165372

T W Korner Fourier analysis Cambridge University Press Cambridge secondedition 1989 ISBN 0-521-38991-7

Cornelius Lanczos Discourse on Fourier series Hafner Publishing Co NewYork 1966

Michel Ledoux The concentration of measure phenomenon volume 89 of Mathe-matical Surveys and Monographs American Mathematical Society ProvidenceRI 2001 ISBN 0-8218-2864-9

Hugh L Montgomery Early Fourier analysis volume 22 of Pure and AppliedUndergraduate Texts American Mathematical Society Providence RI 2014ISBN 978-1-4704-1560-0

Klaus Roth Sur quelques ensembles drsquoentiers C R Acad Sci Paris 234388ndash390 1952 ISSN 0001-4036

K Soundararajan Additive combinatorics Winter 2007 httpmath

stanfordedu~ksoundNotespdf 2010

Elias M Stein and Rami Shakarchi Fourier analysis volume 1 of PrincetonLectures in Analysis Princeton University Press Princeton NJ 2003 ISBN0-691-11384-X An introduction

48